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QJEEAdv22Ma81f6: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma81f6 Solution: Consider the parabola . Let be the focus of the parabola. A…

Consider the parabola y2=4xy^{2}=4 x . Let SS be the focus of the parabola. A pair of tangents drawn to the parabola from the point P=(−2,1)P=(-2,1) meet the parabola at P1P_{1} and P2P_{2} . Let Q1Q_{1} and Q2Q_{2} be points on the lines SP1S P_{1} and SP2S P_{2} respectively such that PQ1P Q_{1} is perpendicular to SP1S P_{1} and PQ2P Q_{2} is perpendicular to SP2S P_{2} . Then, which of the following is/are TRUE?

Multiple Choice Options

  • A SQ1=2S Q_{1}=2
  • B Q1Q2=3105Q_{1} Q_{2}=\frac{3 \sqrt{10}}{5}
  • C PQ1=3P Q_{1}=3
  • D SQ2=1S Q_{2}=1

Step-by-Step Verified Solution

Let P1(t2,2t)\mathrm{P}_1\left(t^2, 2 t\right) then tangent at P1\mathrm{P}_1 will be

ty=x+t2t y=x+t^2

JEE Advanced 2022 Paper 1 Online Mathematics - Parabola Question 5 English Explanation
Since, it passes through (−2,1)(-2,1)

amp;⇒t2−t−2=0amp;⇒t=2,−1\begin{aligned}& \Rightarrow t^2-t-2 =0 \\\\& \Rightarrow t = 2,-1\end{aligned}

So, we get point P1(4,4)\mathrm{P}_1(4, 4) and P2(1,−2)\mathrm{P}_2(1, -2)

Now finding the equation of SP1:4x−3y−4=0\mathrm{SP}_1: 4 x-3 y-4=0

And equation of SP2:x−1=0\mathrm{SP}_2: x-1=0

Now finding the point by foot of point on line formula,

Q1:x1+24=y1−1−3=−(−8−3−4)25=35\mathrm{Q}_1: \frac{x_1+2}{4}=\frac{y_1-1}{-3}=\frac{-(-8-3-4)}{25}=\frac{3}{5}

We get x1=25,y1=−45x_1=\frac{2}{5}, y_1=\frac{-4}{5} and Q2=(1,1)Q_2=(1,1)

Now using the distance formula we get,

amp;SQ1=(1−25)2+(45)2=1amp;Q1Q2=925+8125=3105amp;PQ1=14425+8125=3amp;SQ2=1\begin{aligned}& \mathrm{SQ}_1 =\sqrt{\left(1-\frac{2}{5}\right)^2+\left(\frac{4}{5}\right)^2}=1 \\\\& \mathrm{Q}_1 \mathrm{Q}_2 =\sqrt{\frac{9}{25}+\frac{81}{25}}=\frac{3 \sqrt{10}}{5} \\\\& \mathrm{PQ}_1 =\sqrt{\frac{144}{25}+\frac{81}{25}}=3 \\\\& \mathrm{SQ}_2 =1\end{aligned}

Hence, option (B, C, D) are correct.

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