JEE Mathematics 2023 Question with Solution

Given: The curve is $x^2 + 2x - 4y + 9 = 0$ and the point is $P(1,3)$.

Find: The value of $(AB)^2$.

First rewrite the curve in standard parabola form:

So this is a parabola with vertex $(-1,2)$.

The tangent to $(x+1)^2 = 4(y-2)$ at $P(1,3)$ is obtained using the tangent form for a translated parabola:

To find point $A$, put $x=0$ because the tangent meets the $y$-axis there:

Hence, $A=(0,2)$.

The line through $P(1,3)$ parallel to $x-3y=6$ has the form

Using $P(1,3)$:

So the required line is

or equivalently,

Now this line meets the parabola $y^2 = 4x$ at $B$. Substitute $x = 3y - 8$ into $y^2 = 4x$:

Therefore, $y=4$ or $y=8$.

Corresponding points are:

• If $y=4$, then $x=3(4)-8=4$, so the point is $(4,4)$.
• If $y=8$, then $x=3(8)-8=16$, so the point is $(16,8)$.

Since $B$ lies on the line $2x - 3y = 8$, test both points:

Hence, $B=(16,8)$.

Now compute the square of the distance between $A(0,2)$ and $B(16,8)$:

Therefore, the value of $(AB)^2$ is $292$.

Given: The tangent point is $P(1,3)$ and the auxiliary line through $P$ is parallel to $x-3y=6$.

Find: The correct intersection point $B$ and then $(AB)^2$.

The tangent to the first parabola gives point $A=(0,2)$. Next, the parallel line through $P$ is

This intersects $y^2 = 4x$ at two points.

Substitute $x=3y-8$ into $y^2=4x$:

So the intersection points are $(4,4)$ and $(16,8)$.

Now use the condition that $B$ lies on $2x-3y=8$. Only $(16,8)$ satisfies that condition, so

Finally,

Therefore, the required value is $292$.