Question 13 - Jee advanced Math 2022 P1 Questions with Solutions

Consider the parabola \(y^{2} = 4x\). Let \(S\) be the focus of the parabola. A pair of tangents drawn to the parabola from the point \(P(-2, 1)\) meet the parabola at \(P_{1}\) and \(P_{2}\). Let \(Q_{1}\) and \(Q_{2}\) be points on the lines \(SP_{1}\) and \(SP_{2}\) respectively such that \(PQ_{1}\) is perpendicular to \(SP_{1}\) and \(PQ_{2}\) is perpendicular to \(SP_{2}\). Then, which of the following is/are TRUE?

A) \(SQ_{1} = 2\)

B) \(Q_{2}Q_{1} = \frac{3\sqrt{10}}{5}\)

C)  \(PQ_{1} = 3\)

D) \(SQ_{2} = 1\)

Sol : 

\(y^{2} = 4x\) is a standard parabola \(y^{2} = 4ax\) with the vertex at the origin.

\(\implies 4a = 4 \implies a = 1(> 0)\)

So this is a parabola which opens to the right(since \(a > 0\)) in the Cartesian plane.

\(\implies\) Focus of the parabola is S(1, 0). 

\(PP_{1}\) and \(PP_{2}\) are tangents to the parabola from \(P(-2, 1)\). And \(PQ_{1}\), \(PQ_{2}\) are perpendiculars to \(SP_{1}\) and \(SP_{2}\). 

We are looking for the coordinates of \(P_{1}, P_{2}, Q_{1}, Q_{2}\) to find the lengths in the options.

Using chain rule of derivatives on \(y^{2} = 4x\),

\((2y)\frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}, y \neq 0\)

Using Point-Slope form, the combined equation of tangents is: 

\(y - 1 = \frac{2}{y}(x + 2)\)

\(\implies y^{2} - y = 2x + 4\)

To find the coordinates of \(P_{1}\) and \(P_{2}\), substitute \(x = \frac{y^{2}}{4}\) in the above equation:

\(\implies y^{2} - y = \frac{y^{2}}{2} + 4\)

\(\implies 2y^{2} - 2y = y^{2} + 8\)

\(\implies y^{2} - 2y - 8 = 0\)

\(\implies y = 4\) and \(y = -2\)

\(\implies x = \frac{4^2}{4}\) and \(x = \frac{4}{4}\)

\(\implies x = 4\) and \(x = 1\)

\(\implies\) coordinates of \(P_{1}\) and \(P_{2}\) are \((4, 4)\) and \((1, -2)\)

\(Q_{1}\) is a point that lies on the lines \(SP_{1}\) and \(PQ_{1}\). So solving the pair of equations of these two lines will give us the coordinates of \(Q_{1}\).

Slope of \(SP_{1} = \frac{0 - 4}{1 - 4} = \frac{4}{3}\)

Equation of \(SP_{1}\) using Point-Slope form is :

\(y - 0 = \frac{4}{3}(x - 1)\)

\(\implies \frac{4}{3}x - y - \frac{4}{3} = 0\)….(1)

Line \(PQ_{1}\) is perpendicular to \(SP_{1}\)

\(\implies\) Slope of \(PQ_{1} = -\frac{3}{4}\)

Equation of \(PQ_{1}\) using Point-Slope form is :

\(y - 1 = -\frac{3}{4}(x + 2)\)

\(\implies \frac{3}{4}x + y + \frac{1}{2} = 0\)….(2)

Adding (1) and (2),

\(\frac{4}{3}x + \frac{3}{4}x - \frac{5}{6} = 0\)

\(\implies \frac{25x}{12} = \frac{5}{6}\)

\(\implies x = \frac{10}{25} = \frac{2}{5}\)

\(\implies y = \frac{4}{3}(\frac{2}{5} - 1)\)

         \( = -\frac{4}{5}\)

So the coordinates of \(Q_{1}\) are \((\frac{2}{5}, -\frac{4}{5})\).

In line \(SP_{2}\), x-coordinate of both the points S and \(P_{2}\) is 1. So this line is parallel to Y-axis, and its equation is \(x = 1\).

\(\implies PQ_{2}\) is parallel to X-axis, and its equation is \(y = 1\).

\(\implies\) coordinates of \(Q_{2}\) are \((1, 1)\).

Let’s check the options now.

A) \(SQ_{1} = 2\)

   \(\implies \sqrt{(-\frac{4}{5} - 0)^2 + (\frac{2}{5} - 1)^2} = 2\)

  \(\implies \sqrt{\frac{16 + 9}{25}} = 2\)

\(\implies 1 = 2\)

which is not true

B) \(Q_{2}Q_{1} = \frac{3\sqrt{10}}{5}\)

   \(\implies \sqrt{(-\frac{4}{5} - 1)^2 + (\frac{2}{5} - 1)^2} = \frac{3\sqrt{10}}{5}\)

  \(\implies \sqrt{\frac{81 + 9}{25}} = \frac{3\sqrt{10}}{5}\)

\(\implies \frac{\sqrt{90}}{5} = \frac{3\sqrt{10}}{5}\)

\(\implies \frac{3\sqrt{10}}{5} = \frac{3\sqrt{10}}{5}\)

which is true

C) \(PQ_{1} = 3\)   

\(\implies \sqrt{(-\frac{4}{5} - 1)^2 + (\frac{2}{5} + 2)^2} = 3\)

  \(\implies \sqrt{\frac{81 + 144}{25}} = 3\)

\(\implies \sqrt{\frac{225}{25}} = 3\)

\(\implies 3 = 3\)

which is true

D) \(SQ_{2} = 1\)

   \(\implies \sqrt{(1 - 0)^2 + (1 - 1)^2} = 1\)

  \(\implies \sqrt{1} = 1\)

\(\implies 1 = 1\)

which is true

\(\implies\) Options B, C and D are True.

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