datavorous/entrance-exam-dataset huggingface dataset:
The equation of the hyperbola referred to its axes as axes of coordinate and whose distance between the foci is 16 and eccentricity is $\sqrt{2}$, is

Options:
A $x^2-y^2=16$
B $x^2-y^2=32$
C $x^2-2 y^2=16$
D $y^2-x^2=16$

The correct answer is:
$x^2-y^2=32$

Solution:
2ae $=16, e=\sqrt{2} \Rightarrow a=4 \sqrt{2}$ and $b=4 \sqrt{2}$
$\therefore$ equation is $\frac{x^2}{(4 \sqrt{2})^2}-\frac{y^2}{(4 \sqrt{2})^2}=1 \quad \Rightarrow x^2-y^2=32$.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4 x+3 y-7=0$ and $x-2 y-1=0$ is

Options:
A $4 x^2+5 x y-6 y^2-11 x+11 y+50=0$
B $4 x^2+5 x y-6 y^2-11 x+11 y-43=0$
C $4 x^2-5 x y-6 y^2-11 x+11 y+57=0$
D $x^2-5 x y-y^2-11 x+11 y-43=0$

The correct answer is:
$4 x^2-5 x y-6 y^2-11 x+11 y+57=0$

Solution:
Since, the equation of the hyperbola differs from that of the joint equations of the tangents by a constant, therefore the equation of the hyperbola will be of the form
$(4 x+3 y-7)(x-2 y-1)+k=0$ ...(i)
$\therefore$ Since, the hyperbola passes through the point $(2,3)$.
$\Rightarrow \quad(8+9-7)(2-6-1)+k=0$
$\Rightarrow \quad(10)(-5)+k=0,-50+k=0$
$\Rightarrow \quad k=50$
Hence, from Eq. (i)
$(4 x+3 y-7)(x-2 y-1)+50=0$
$4 x^2+3 x y-7 x-8 x y-6 y^2+14 y-4 x$ $-3 y+7+50=0$
$4 x^2-5 x y-6 y^2-11 x+11 y+57=0$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13 , is

Options:
A $25 x^2-144 y^2=900$
B $144 x^2-25 y^2=900$
C $144 x^2+25 y^2=900$
D $25 x^2+144 y^2=900$

The correct answer is:
$25 x^2-144 y^2=900$

Solution:
Conjugate axis is 5 and distance between foci $=13 \Rightarrow 2 b=5$ and $2 a e=13$. Now, also we know for hyperbola
$\begin{aligned}
& b^2=a^2\left(e^2-1\right) \Rightarrow \frac{25}{4}=\frac{(13)^2}{4 e^2}\left(e^2-1\right) \\
& \Rightarrow \frac{25}{4}=\frac{169}{4}-\frac{169}{4 e^2} \text { or } e^2=\frac{169}{144} \Rightarrow e=\frac{13}{12} \\
& \text { or } a=6, b=\frac{5}{2} \text { or hyperbola is } \frac{x^2}{36}-\frac{y^2}{25 / 4}=1 \\
& \Rightarrow 25 x^2-144 y^2=900 .
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the hyperbola with vertices $(3,0),(-3,0)$ and semi-latus rectum 4 is given by :

Options:
A $4 x^{2}-3 y^{2}+36=0$
B $4 x^{2}-3 y^{2}+12=0$
C $4 x^{2}-3 y^{2}-36=0$
D $4 x^{2}+3 y^{2}-25=0$

The correct answer is:
$4 x^{2}-3 y^{2}-36=0$

Solution:
We have $a=3$ and $\frac{b^{2}}{a}=4 \Rightarrow b^{2}=12$
Hence, the equation of the hyperbola is
$$
\begin{array}{l}
\frac{x^{2}}{9}-\frac{y^{2}}{12}=1 \\
\Rightarrow 4 x^{2}-3 y^{2}=36 \Rightarrow 4 x^{2}-3 y^{2}-36=0
\end{array}
$$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the image of line y=x with respect to the line mirror 2x-y=1 is

Options:
A y=7x-5
B y=7x-6
C y=3x-7
D y=6x-5

The correct answer is:
y=7x-6

Solution:
Image of 0,0 w.r.t L lies on L1
Image of 0,0 w.r.t L is
x-02=y-0-1=-2-15
P'≡x,y=45,-25
Equation of L'1 which passes through 1,1 and 45,-25 is
y-1=-25-145-1x-1=-7-1x-1
y=7x-6

Topic: Mathematics, Straight Lines, JEE Main
The equation of the line bisecting the line segment joining the points $(a, b)$ and ${ }^{\left(a^{\prime}, b^{\prime}\right)}$ at right angle, is

Options:
A $2\left(a-a^{\prime}\right) x+2\left(b-b^{\prime}\right) y=a^2+b^2-a^{\prime 2}-b^{\prime 2}$
B $\left(a-a^{\prime}\right) x+\left(b-b^{\prime}\right) y=a^2+b^2-a^{\prime 2}-b^{\prime 2}$
C $2\left(a-a^{\prime}\right) x+2\left(b-b^{\prime}\right) y=a^{\prime 2}+b^{\prime 2}-a^2-b^2$
D None of these

The correct answer is:
$2\left(a-a^{\prime}\right) x+2\left(b-b^{\prime}\right) y=a^2+b^2-a^{\prime 2}-b^{\prime 2}$

Solution:
$\begin{gathered}
m=\frac{-1}{\frac{b^{\prime}-b}{a^{\prime}-a}}=\frac{a^{\prime}-a}{b-b^{\prime}} \\
\text {. Mid point is }\left(\frac{a+a^{\prime}}{2}, \frac{b+b^{\prime}}{2}\right) \\
\text {Therefore equation of line is } y-\left(\frac{b+b^{\prime}}{2}\right)=\frac{a^{\prime}-a}{b-b^{\prime}}\left(x-\frac{a+a^{\prime}}{2}\right) \\
\Rightarrow 2\left(b-b^{\prime}\right) y+2\left(a-a^{\prime}\right) x-b^2+b^{\prime 2}-a^2+a^{\prime 2}=0 .
\end{gathered}$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the line joining the origin to the point $(-4,5)$, is

Options:
A $5 x+4 y=0$
B $3 x+4 y=2$
C $5 x-4 y=0$
D $4 x-5 y=0$

The correct answer is:
$5 x+4 y=0$

Solution:
$m=\frac{5}{-4}$. Therefore the line is $5 x+4 y=0$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the line passing through -4, 3, 1, parallel to the plane x+2y-z-5=0 and intersecting the line x + 1-3=y-32=z-2-1 is

Options:
A x+43=y-3-1=z-11
B x+41=y-31=z-13
C x+4-1=y-31=z-11
D x-42=y+31=z+14

The correct answer is:
x+43=y-3-1=z-11

Solution:
A variable point on line x + 1-3=y-32=z-2-1 is -3t-1, 2t+3,-t+2
∴DR of variable point & -4, 3, 1 is
-3t+3, 2t, -t+1
Since line is parallel to x+2y-z-5=0
∴-3t+3+4t+t-1=0
⇒t=-1
∴DR of line is 6,-2, 2 or 3, -1, 1
∴ Equation of line is x+43=y-3-1=z-11

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The equation of the line passing through $(4,-6)$ and makes an angle $45^{\circ}$ with positive $x$-axis, is

Options:
A $x-y-10=0$
B $x-2 y-16=0$
C $x-3 y-22=0$
D None of these

The correct answer is:
$x-y-10=0$

Solution:
The required equation is $y+6=\tan 45^{\circ}(x-4) \Rightarrow x-y-10=0$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the line passing through the point of intersection of the lines $x-3 y+2=0$ and $2 x+5 y-7=0$ and perpendicular to the line $3 x+2 y+5=0$ is :

Options:
A $2 x-3 y+1=0$
B $6 x-9 y+11=0$
C $2 x-3 y+5=0$
D $3 x-2 y+1=0$

The correct answer is:
$2 x-3 y+1=0$

Solution:
The equation of lines are
$x-3 y+2=0$ ...(i)
and $2 x+5 y-7=0$ ...(ii)
On solving Eqs. (i) and (ii), we get
$x=1, y=1$
$\therefore$ The co-ordinates of a point of intersection of given lines are $(1,1)$.
The equation of line perpendicular to $3 x+2 y+5=0$ is $2 x-3 y+\lambda=0$.
Which passes through $(1,1)$.
$\therefore \quad 2-3+\lambda=0$
$\Rightarrow \quad \lambda=1$
$\therefore$ Required equation of line is
$2 x-3 y+1=0$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the line through the point 0,1,2 and perpendicular to the line x-12=y+13=z-1-2 is :

Options:
A x3=y-1-4=z-23
B x3=y-14=z-23
C x-3=y-14=z-23
D x3=y-14=z-2-3

The correct answer is:
x-3=y-14=z-23

Solution:
x-12=y+13=z-1-2=r

⇒ Px,y,z=2r+1,3r-1,-2r+1

Since, QP→⊥2i^+3j^-2k^

⇒ 4r+2+9r-6+4r+2=0

⇒ r=217

⇒ P2117,-1117,1317

⇒ PQ→=21i^-28j^-21k^17

So, QP→:x-3=y-14=z-23

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The equation of the line which cuts off an intercept 3 units on $O X$ and an intercept -2 unit on $O Y$, is

Options:
A $\frac{x}{3}-\frac{y}{2}=1$
B $\frac{x}{3}+\frac{y}{2}=1$
C $\frac{x}{2}+\frac{y}{3}=1$
D $\frac{x}{2}-\frac{y}{3}=1$

The correct answer is:
$\frac{x}{3}-\frac{y}{2}=1$

Solution:
Here intercept on $x$-axis is 3 and intercept on $y$-axis is -2 . So using intercept form of the equation of line,the required line is $\frac{x}{3}-\frac{y}{2}=1$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the line which intersect each of the two lines 2x+y-1=0=x-2y+3z and 3x-y+z+2=0=4x+5y-2z-3=0 and is parallel to x1=y2=z3 is

Options:
A 4x+7y-6z-1=0=2x-7y+4z+3
B 4x+7y-6z-4=0=2x-7y+4z+2
C 4x+7y-6z-3=0=2x-7y+4z+7
D 4x+7y-6z+7=0=2x-7y+4z-3

The correct answer is:
4x+7y-6z-3=0=2x-7y+4z+7

Solution:
Equation of the first plane is 2x+y-1+λx-2y+3z=0
⇒x2+λ+y1-2λ+3λz-1=0
Equation of the second plane is 3x-y+z+2+u4x+5y-2z-3=0
⇒x3+4u+y-1+5u+z1-2u+2-3u=0
Required line must lie in both planes and parallel to the line x1=y2=z3
⇒2+λ1+1-2λ2+3λ3=0⇒λ=-23
⇒3+4u1+-1+5u2+1-2u3=0
⇒8u+4=0⇒u=-12
Equation of both planes is 4x+7y-6z-3=0 and 2x-7y+4z+7=0
So the equation of the required line is
4x+7y-6z-3=0=2x-7y+4z+7

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the line whose slope is 3 and which cuts off an intercept 3 from the positive $x-$ axis is

Options:
A $y=3 x-9$
B $y=3 x+3$
C $y=3 x+9$
D None of these

The correct answer is:
$y=3 x-9$

Solution:
Given: Slope \(m=3\)
Since the line cuts off an intercept 3 from the positive \(x\) axis.
So it passes through \((3,0)\).
Equation of line passing through (\(x_1, y_1\)) is given by,
\(\begin{aligned}
& \left(y-y_1\right)=m\left(x-x_1\right) \\
& \Rightarrow y-0=3(x-3) \\
& \Rightarrow y=3 x-9
\end{aligned}\)
Hence Option (A) is the correct answer.

Topic: Mathematics, Straight Lines, JEE Main
The equation of the lines which passes through the point $(3,-2)$ and are inclined at $60^{\circ}$ to the line $\sqrt{3} x+y=1$

Options:
A $y+2=0, \sqrt{3} x-y-2-3 \sqrt{3}=0$
B $x-2=0, \sqrt{3} x-y+2+3 \sqrt{3}=0$
C $\sqrt{3} x-y-2-3 \sqrt{3}=0$
D None of these

The correct answer is:
$y+2=0, \sqrt{3} x-y-2-3 \sqrt{3}=0$

Solution:
The equation of any straight line passing through
$(3,-2) \text { is } y+2=m(x-3)$
The slope of the given line is $-\sqrt{3}$
So,
$\tan 60^{\circ}= \pm \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})}$
On solving, we get $m=0$ or $\sqrt{3}$
Putting the values of $m$ in (i), the required equation of lines are $y+2=0$ and
$\sqrt{3} x-y=2+3 \sqrt{3}$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the locus of the foot of perpendicular drawn from 5,6 on the family of lines x-2+λy-3=0 (where λ∈R) is

Options:
A x-1x-3+y-2y-6=0
B x-5x-6+y-2y-3=0
C x-2x-5+y-3y-6=0
D x+2x+5+y+3y+6=0

The correct answer is:
x-2x-5+y-3y-6=0

Solution:
Let A=5,6 and the point of concurrency of the family of lines x-2+λy-3=0 is 2,3=B and foot of the perpendicular from A to the family of lines is P=h,k
Now, PA is perpendicular to PB
⇒slope of PA×slope of PB=-1
⇒k-6h-5×k-3h-2=-1⇒k-6k-3=-h-5h-2
⇒ locus is x-2x-5+y-3y-6=0

Topic: Mathematics, Straight Lines, JEE Main
The equation of the normal at the point $\left(\frac{a}{4}, a\right)$ to the parabola $y^2=4 a x$, is

Options:
A $4 x+8 y+9 a=0$
B $4 x+8 y-9 a=0$
C $4 x+y-a=0$
D $4 x-y+a=0$

The correct answer is:
$4 x+8 y-9 a=0$

Solution:
$y-a=\frac{-a}{2 a}\left(x-\frac{a}{4}\right)$
$\Rightarrow \quad 2 y+x=2 a+\frac{a}{4}=\frac{9 a}{4}$ $\Rightarrow \quad 2 y+x-\frac{9 a}{4}=0$
i.e., $4 x+8 y-9 a=0$

Topic: Mathematics, Parabola, JEE Main
The equation of the normal at the point $(6,4)$ on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=3$, is

Options:
A $3 x+8 y=50$
B $3 x-8 y=50$
C $8 x+3 y=50$
D $8 x-3 y=50$

The correct answer is:
$3 x+8 y=50$

Solution:
Equation of normal at any point $\left(x_1, y_1\right)$ on hyperbola is, $\frac{a^2\left(x-x_1\right)}{x_1}=\frac{b^2\left(y-y_1\right)}{-y_1}$
Here, $a^2=267, b^2=48$ and $\left(x_1, y_1\right)=(6,4)$
$\begin{aligned}
& \therefore \frac{27(x-6)}{6}=-\frac{48(y-4)}{4} \Rightarrow 3(x-6)=-8(y-4) \\
& \Rightarrow 3 x+8 y=50 .
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the normal at the point $(a \sec \theta, b \tan \theta)$ of the curve $b^2 x^2-a^2 y^2=a^2 b^2$ is

Options:
A $\frac{a x}{\cos \theta}+\frac{b y}{\sin \theta}=a^2+b^2$
B $\frac{a x}{\tan \theta}+\frac{b y}{\sec \theta}=a^2+b^2$
C $\frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}=a^2+b^2$
D $\frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}=a^2-b^2$

The correct answer is:
$\frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}=a^2+b^2$

Solution:
Equation of normal to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $(a \sec \theta, b \tan \theta)$ is $\frac{a^2 x}{a \sec \theta}+\frac{b^2 y}{b \tan \theta}=a^2+b^2$. (Direct formula based)

Topic: Mathematics, Hyperbola, JEE Main
The equation of the normal to the circle $x^2+y^2=9$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ is

Options:
A $x+y=0$
B $x-y=\frac{\sqrt{2}}{3}$
C $x-y=0$
D None of these

The correct answer is:
$x-y=0$

Solution:
We know that the equation of normal to the circle $x^2+y^2=a^2$ at the point $\left(x_1, y_1\right)$ is $\frac{x}{x_1}-\frac{y}{y_1}=0$
Therefore,$\frac{x}{1 / \sqrt{2}}-\frac{y}{1 / \sqrt{2}}=0 \Rightarrow x-y=0$.

Topic: Mathematics, Circle, JEE Main
The equation of the normal to the curve $3 x^2-y^2=8$, which is parallel to the line $x+3 y=10$, is

Options:
A $x+3 y+6=0$
B $x+3 y-3=0$
C $x+3 y+8=0$
D $x+3 y-4=0$

The correct answer is:
$x+3 y+8=0$

Solution:
$$
3 x^2-y^2=8
$$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& 6 x-2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x}{y}
\end{aligned}
$$
$\therefore \quad$ Slope of the tangent to the curve is $\frac{3 x}{y}$.
$\therefore \quad$ Slope of the normal is $\frac{-y}{3 x}$.
It is parallel to line $x+3 y=10 \Rightarrow$ slope $=-\frac{1}{3}$
$\therefore \quad \frac{-y}{3 x}=\frac{-1}{3} \Rightarrow x=y$
$\therefore \quad$ When $x=y$, equation of the curve becomes
$\therefore \quad 3 x^2-x^2=8$
$\therefore \quad x^2=4$
$\therefore \quad x=2,-2 \Rightarrow y=2,-2$
$\therefore \quad(2,2)$ and $(-2,-2)$ are the points of contact of the normal and the curve.
$\therefore \quad$ Equations are $(y-2)=\frac{-1}{3}(x-2)$ or
$$
\begin{aligned}
& (y+2)=\frac{-1}{3}(x+2) \\
& \text { i.e., } x+3 y-8=0 \text { or } x+3 y+8=0
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of the normal to the curve y=1+x2y+cos2sin-1x, at x=0 is

Options:
A y+4x=2
B y=4x+2
C x + 4y = 8
D 2y + x = 4

The correct answer is:
x + 4y = 8

Solution:
y=1+x2y+cos2sin-1x
Now at x=0, y=2

Let sin-1x=t ⇒sint=x

⇒cos2sin-1x=cos2t=1-sin2t=1-x2

y=1+x2y+1-x2

y=e2yln1+x+1-x2

dydx=e2yln(1+x)2y1+x+ln(1+x)·2dydx-2x

Now at x=0, dydx=2×21+0+0

dydx=4
y-2=-14x-0
4y-8=-x
x+4y=8

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The equation of the normal to the curve $y=\cosh x$ drawn at the point nearest to the origin is

Options:
A $y=0$
B $x=1$
C $x=0$
D $y=1$

The correct answer is:
$x=0$

Solution:
$\because y=\cos h x$
$\Rightarrow y^{\prime}=\sin \mathrm{hx}$
To find the nearest point to the origin, we can set the derivative equal to zero.
$$
\therefore \sin \mathrm{h}(\mathrm{x})=0
$$
The only solution of this equation is $x=0$
At $x=0, y=\operatorname{cash}(0)=1$
The slope of normal is
$$
\mathrm{m}=\left(-\frac{1}{\mathrm{y}^1}\right)_{(0,1)}=\left(-\frac{1}{\sinh x}\right)_{(0,1)}
$$
$$
\mathrm{m}=\frac{-1}{0}
$$
Now, the equation of the normal is
$$
(y-1)=-\frac{1}{0}(x-0) \Rightarrow(x-0)=0 \Rightarrow x=0
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of the normal to the curve $y=\sin \frac{\pi x}{2}$ at $(1,1)$ is

Options:
A $y=1$
B $x=1$
C $y=x$
D $y-1=\frac{-2}{\pi}(x-1)$

The correct answer is:
$x=1$

Solution:
$y=\sin \frac{\pi x}{2} \Rightarrow \frac{d y}{d x}=\frac{\pi}{2} \cos \frac{\pi}{2} x \Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=0$
$\therefore$ Equation of normal is $\quad y-1=\frac{1}{0}(x-1)$
$\Rightarrow x=1 \text {. }$

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the positive end of latus rectum in the first quadrant

Options:
A $x+e y+e^2 a=0$
B $x-e y-e^3 a=0$
C $x-e y-e^2 a=0$
D None of these

The correct answer is:
$x-e y-e^3 a=0$

Solution:
The correct option is B
$x-e y-e^3 a=0$
The positive end of latus rectum is ( $a e, \frac{b^2}{a}$ )
$\begin{aligned} & \mathrm{a} \cos \theta=\mathrm{ae} \therefore \cos \theta=\mathrm{e} \\ & \mathrm{b} \sin \theta=\frac{\mathrm{b}^2}{\mathrm{a}} \text { and } \sin \theta=\frac{\mathrm{b}}{\mathrm{b}}\end{aligned}$
Equation of normal is $\frac{a x}{\cos \theta}-\frac{\text { by }}{\sin \theta}=a^2-b^2$
$\begin{aligned} & \therefore \frac{a x}{e}-\frac{b y}{b} \times a=a^2\left(e^2\right) \\ & \text { or } x-e y=a e^3 \\ & \therefore x-e y-e^3 a=0\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The equation of the normal to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ at $(-4,0)$ is

Options:
A $y=0$
B $y=x$
C $x=0$
D $x=-y$

The correct answer is:
$y=0$

Solution:
$\frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow \frac{2 x}{16}-\frac{2 y}{9} \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{2 x \times 9}{16 \times 2 y}=\frac{9}{16} \frac{x}{y} \Rightarrow\left(\frac{-d x}{d y}\right)_{(-4,0)}=\frac{-16}{9} \frac{y}{x}=0$
Hence, equation of normal
$\Rightarrow(y-0)=0(x+4) \Rightarrow y=0 \text {. }$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the normal to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ at the point $(8,3 \sqrt{3})$ is

Options:
A $\sqrt{3} x+2 y=25$
B $x+y=25$
C $y+2 x=25$
D $2 x+\sqrt{3} y=25$

The correct answer is:
$2 x+\sqrt{3} y=25$

Solution:
Applying the formula, the required normal is
$\frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9 \text { i.e., } 2 x+\sqrt{3} y=25$
Trick : This is the only equation among the given options at which the point $(8,3 \sqrt{3})$ is located.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the normal to the parabola, $x^2=8 y$ at $x=4$ is

Options:
A
$x+2 y=0$

B
$x+y=2$

C
$x-2 y=0$

D
$x+y=6$

The correct answer is:
$x+y=6$

Solution:
$x^2=8 y$
When, $x=4$, then $y=2$
Now $\left.\frac{d y}{d x}=\frac{2 x}{8}=\frac{x}{4}, \frac{d y}{d x}\right]_{x=4}=1$
Slope of normal $=-\frac{1}{\frac{d y}{d x}}=-1$
Euqation of normal at $x=4$ is $y-2=-1(x-4)$
$\Rightarrow y=-x+4+2=-x+6$
$\Rightarrow x+y=6$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2012 (19 May Online)
The equation of the parabola whose vertex and focus lies on the $x$-axis at distance $a$ and $a^{\prime}$ from the origin, is

Options:
A $y^2=4\left(a^{\prime}-a\right)(x-a)$
B $y^2=4\left(a^{\prime}-a\right)(x+a)$
C $y^2=4\left(a^{\prime}+a\right)(x-a)$
D $y^2=4\left(a^{\prime}+a\right)(x+a)$

The correct answer is:
$y^2=4\left(a^{\prime}-a\right)(x-a)$

Solution:
Equation will be of the form $y^2=4 A(x-a)$, where $A=\left(a^{\prime}-a\right)$ or $y^2=4\left(a^{\prime}-a\right)(x-a)$.

Topic: Mathematics, Parabola, JEE Main
The equation of the parabola with focus $(0,0)$ and directrix $x+y=4$ is

Options:
A $x^2+y^2-2 x y+8 x+8 y-16=0$
B $x^2+y^2-2 x y+8 x+8 y=0$
C $x^2+y^2+8 x+8 y-16=0$
D $x^2-y^2+8 x+8 y-16=0$

The correct answer is:
$x^2+y^2-2 x y+8 x+8 y-16=0$

Solution:
Given focus of parabola is $S(0,0)$.
Equation of directrix is $x+y=4$
Let $P(x, y)$ be any point on the parabola.
Now, $S P^2=P M^2$
$\Rightarrow \quad(x-0)^2+(y-0)^2=\left[\frac{x+y-4}{\sqrt{1+1}}\right]^2$
$\Rightarrow \quad x^2+y^2=\frac{(x+y-4)^2}{2}$
$\Rightarrow \quad 2 x^2+2 y^2=x^2+y^2+16+2 x y-8 x-8 y$
$\therefore \quad x^2+y^2-2 x y+8 x+8 y-16=0$

Topic: Mathematics, Parabola, JEE Main
The equation of the parabola with focus $(3,0)$ and the directixx $x+3=0$ is

Options:
A $y^2=3 x$
B $y^2=2 x$
C $y^2=12 x$
D $y^2=6 x$

The correct answer is:
$y^2=12 x$

Solution:
$S P^2=P M^2$
$\Rightarrow \quad(x-3)^2+y^2=\left|\frac{x+3}{\sqrt{1}}\right|^2$
$\Rightarrow x^2+9-6 x+y^2=x^2+9+6 x$
$\Rightarrow y^2=12 x$.

Topic: Mathematics, Parabola, JEE Main
The equation of the parabola with focus $(a, b)$ and directrix $\frac{x}{a}+\frac{y}{b}=1$ is given by

Options:
A $(a x-b y)^2-2 a^3 x-2 b^3 y+a^4+a^2 b^2+b^4=0$
B $(a x+b y)^2-2 a^3 x-2 b^3 y-a^4+a^2 b^2-b^4=0$
C $(a x-b y)^2+a^4+b^4-2 a^3 x=0$
D $(a x-b y)^2-2 a^3 x=0$

The correct answer is:
$(a x-b y)^2-2 a^3 x-2 b^3 y+a^4+a^2 b^2+b^4=0$

Solution:
$(x-a)^2+(y-b)^2$ $=\left(\frac{b x+a y-a b}{\sqrt{a^2+b^2}}\right)^2$
On solving we get
$(a x-b y)^2-2 a^3 x-2 b^3 y+a^4+a^2 b^2+b^4=0$

Topic: Mathematics, Parabola, JEE Main
The equation of the parabola with the focus $(3,0)$ and the directrix $x+3=0$, is.

Options:
A $y^2=3 x$
B $y^2=6 x$
C $y^2=12 x$
D $y^2=2 x$

The correct answer is:
$y^2=12 x$

Solution:
Given that focus is $S(3,0)$, let $P(x, y)$ be any point on the parabola.
$\therefore$ Directrix is, $x+3=0$
Also, $S P^2=P M^2$
$$
\begin{aligned}
& \Rightarrow \quad(x-3)^2+(y-0)^2=\left(\frac{x+3}{\sqrt{1}}\right)^2 \\
& \Rightarrow \quad(x-3)^2+y^2=(x+3)^2 \\
& \Rightarrow \quad y^2=(x+3)^2-(x-3)^2 \\
& \Rightarrow \quad=(x+3+x-3)(x+3-x+3) \\
& \Rightarrow \quad y^2=2 x 6=12 x
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main
The equation of the plane bisecting the line segment joining the points $(2,0,6)$ and $(-6,2,4)$ and perpendicular to it, is

Options:
A $2 x-y+4 z-15=0$
B $4 x-y+3 z-6=0$
C $4 x-y+z+4=0$
D $x-2 y+3 z-11=0$

The correct answer is:
$4 x-y+z+4=0$

Solution:
Let variable point on the plane is $(x, y, z)$.
Mid-point of the line segment joining the points $(2,0,6)$ and $(-6,2,4)=(-2,1,5)$
Direction ratio of the line segment joining the points $(2,0,6)$ and $(-6,2,4)$
$$
\Rightarrow(-6-2),(2-0),(4-6)-8,2,-2=a_1, b_1, c_1
$$
direction ratios of line joining $(x, y, z)$ and $(-2,1,5)$
$$
=(x+2),(y-1),(z-5)=a_2, b_2, c_2
$$
direction ratio's $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are perpendicular so,
$$
a_1 a_2+b_1 b_2+c_1 c_2=0
$$
So, required equation of plane is
$$
\begin{array}{rrrl}
\Rightarrow & -8(x+2)+2(y-1)-2(z-5) & =0 \\
\Rightarrow & -8 x-16+2 y-2-2 z+10 & =0 \\
\Rightarrow & -8 x+2 y-2 z-8 & =0 \\
\Rightarrow & & 4 x-y+z+4 & =0 .
\end{array}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

Options:
A $x+y+z=1$
B $x+y+z=2$
C $x+y+z=0$
D None of these

The correct answer is:
$x+y+z=0$

Solution:
The equation of plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ is
$$
a(x+1)+b(y-3)+c(z+2)=0 ...(i)
$$
where $\quad-3 a+2 b+c=0$ ...(ii)
This passes through $(0,7,-7)$.
$\therefore$
$a+4 b-5 c=0$ ...(iii)
From Eqs.
(ii) and (iii),
or
$$
\begin{aligned}
\frac{a}{-14} &=\frac{b}{-14}=\frac{c}{-14} \\
\frac{a}{1} &=\frac{b}{1}=\frac{c}{1}
\end{aligned}
$$
Thus, the required plane is $x+y+z=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane containing the line of intersection of 2x-5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is

Options:
A 2 x + 6 y + 1 2 z = - 1 3
B 2 x + 6 y + 1 2 z = 1 3
C x + 3 y + 6 z = - 7
D x + 3 y + 6 z = 7

The correct answer is:
x + 3 y + 6 z = 7

Solution:
Any plane containing the line of intersection of 2x-5y+z=3, x+y+4z=5 will be of the form
2 x - 5 y + z - 3 + λ x + y + 4 z - 5 = 0
2 + λ x - 5 - λ y + 1 + 4 λ z - 3 + 5 λ = 0
It is parallel to plane x + 3 y + 6 z = 1
⇒ 2+λ1=λ-53=1+4λ6≠3+5λ1
2 + λ 1 = λ - 5 3 ⇒ 6 + 3 λ = λ - 5
2 λ = - 1 1
λ = - 1 1 2
∴ Required plane is
2x-5y+z-3+-112x+y+4z-5=0
2 2 x - 5 y + z - 3 - 1 1 x + y + 4 z - 5 = 0
4 x - 1 0 y + 2 z - 6 - 1 1 x - 1 1 y - 4 4 z + 5 5 = 0
- 7 x - 2 1 y - 4 2 z + 4 9 = 0
⇒ x + 3 y + 6 z - 7 = 0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2015 (04 Apr)
The equation of the plane containing the line of intersection of the planes \( 2 x-y=0 \) and \( y-3 z=0 \) and perpendicular to the plane \( 4 x+5 y-3 z-8=0 \) is

Options:
A \( 28 x-17 y+9 z=0 \)
B \( 28 x+17 y+9 z=0 \)
C \( 28 x-17 y+9 x=0 \)
D \( 7 x-3 y+z=0 \)

The correct answer is:
\( 28 x-17 y+9 z=0 \)

Solution:
Equation of plane containing the line of intersection of planes is, 2x-y+λ y-3z=0 ......(i)
Also, plane (i) is perpendicular to 4x+5y-3z-8=0
∴ 4 2+5 λ-1-3 -3λ=0
⇒14λ=-3⇒λ=-314
Put the value of λ in (i), we get 28x-17y+9z=0 , which is the required plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through $(1,1,1)$ and $(1,-1,-1)$ and perpendicular to $2 x-y+z+5=0$ is :

Options:
A $2 x+5 y+z+8=0$
B $x+y-z-1=0$
C $2 x+5 y+z+4=0$
D $x-y+z-1=0$

The correct answer is:
$x+y-z-1=0$

Solution:
Let the equation of plane passing through $(1,1,1)$ is
$a(x-1)+b(y-1)+c(z-1)=0$ $\ldots$ (i)
It is also passing through $(1,-1,-1)$.
$\therefore \quad a(0)+b(-2)+c(-2)=0$
or $b+c=0$ $\ldots$ (ii)
Since, Eq. (i) is perpendicular to the plane
$2 x-y+z+5=0$
$\therefore \quad 2 a-b+c=0$ $\ldots$ (iii)
Since, Eqs. (ii) and (iii) are identical
$\therefore \quad \frac{a}{-2-2}=\frac{b}{-4-0}=\frac{c}{4} \Rightarrow \frac{a}{-4}=\frac{b}{-4}=\frac{c}{4}$
$\therefore$ Required equation of plane is
$x+y-z-1=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main