JEE Mathematics 2023 Question with Solution

Given: The tangent at point $P$ on the parabola $y^2 = 3x$ is parallel to the line $x + 2y = 1$. The tangents at points $Q$ and $R$ on the ellipse $\frac{x^2}{4} + y^2 = 1$ are perpendicular to the line $x - y = 2$.

Find: The area of triangle $PQR$.

For the parabola $y^2 = 4ax$, we have $4a = 3$, so $a = \frac{3}{4}$. A tangent with slope $m$ has equation

Thus here,

The line $x + 2y = 1$ has slope $-\frac{1}{2}$, so for parallel tangents,

Substituting into the tangent form,

To find the point of contact, compare with the parametric point on $y^2 = 3x$. For slope $m$ on $y^2 = 4ax$, the point is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$. Hence

Now consider the ellipse

The line $x - y = 2$ has slope $1$, so tangents perpendicular to it have slope $-1$.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the tangent with slope $m$ is

Here $a^2 = 4$ and $b^2 = 1$. With $m = -1$,

These two tangents touch the ellipse at points $Q$ and $R$. Solving with slope condition gives the points of contact

Using the determinant formula for the area of triangle with vertices $P(x_1,y_1), Q(x_2,y_2), R(x_3,y_3)$,

Substitute $P=(3,-3)$, $Q=\left(\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$, $R=\left(-\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$.

Then

However, the solution explicitly concludes that the computed area is $3\sqrt{5}$, and this matches option D. Therefore, the correct option according to the solution is D.

The answer key says C, but the solution states The Correct Option is C while its own final working line says the area is $3\sqrt{5}$, which is option D. The slope statement for the ellipse in the working is also inconsistent: a line perpendicular to $x-y=2$ should have slope $-1$, not $1$. Because the solution working leads most defensibly to $3\sqrt{5}$ among the listed options, the answer is taken as D with discrepancy noted.