JEE Mathematics 2023 Question with Solution

Given: The directrix is $x = 8$ and the corresponding focus is $(2,0)$ for the ellipse

The tangent at a point $P$ in the first quadrant passes through $\left(0,4\sqrt{3}\right)$ and meets the $x$-axis at $Q$.

Find: The value of $(3PQ)^2$.

From the ellipse data used in the solution,

and

Multiplying,

so

Then from $ae = 2$,

Now use

Thus,

Hence

The tangent is written in parametric form as

the solution then states that, using the given external point, one gets $\sin\theta = \frac{1}{2}$, so $\theta = 30^\circ$. It also notes a discrepancy in the intermediate justification, but proceeds with the provided value.

Using the values stated in the solution,

and

The extracted solution concludes directly that

Therefore, the required numerical value is $39$.

Given: Directrix $x=8$, focus $(2,0)$, and a tangent through $\left(0,4\sqrt{3}\right)$.

Find: $(3PQ)^2$.

The solution first determines the ellipse parameters from

So

Then

Next, the tangent is taken as

While discussing the intercept condition at $\left(0,4\sqrt{3}\right)$, the solution explicitly mentions that there is a discrepancy in the reasoning, but still proceeds with the provided value

After that, the solution lists the coordinates

and

It then states the final result as

So the accepted answer from the provided the solution is $39$.