Consider the parabola y 2 = 4 x . 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 o

Question

Consider the parabola y 2 = 4 x . 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 S P 1 and S P 2 respectively such that P Q 1 is perpendicular to S P 1 and P Q 2 is perpendicular to S P 2 . Then, which of the following is/are TRUE?

TestHub · Accepted Answer

On plotting the diagram of the given value we have,

Now let $P_{1} (t^{2}, 2 t)$ then tangent at $P_{1}$ will be $t y = x + t^{2}$

Since it passes through $(- 2, 1)$

$\Rightarrow t^{2} - t - 2 = 0$

$\Rightarrow t = 2, - 1$

So, we get points $P_{1} (4, 4)$ and $P_{2} (1, - 2)$

Now finding the equation of $S P_{1} : 4 x - 3 y - 4 = 0$

And equation of $S P_{2} : x - 1 = 0$

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

$Q_{1} : \frac{x_{1} + 2}{4} = \frac{y_{1} - 1}{- 3} = \frac{- (- 8 - 3 - 4)}{25} = \frac{3}{5}$

We get $x_{1} = \frac{2}{5}, y_{1} = \frac{- 4}{5}$

Similarly for $Q_{2} = (1, 1)$

Now using the distance formula we get, $S Q_{1} = \sqrt{{(1 - \frac{2}{5})}^{2} + {(\frac{4}{5})}^{2}} = 1$

$Q_{1} Q_{2} = \sqrt{\frac{9}{25} + \frac{81}{25}} = \sqrt{\frac{90}{25}} = \frac{3 \sqrt{10}}{5}$

$P Q_{1} = \sqrt{\frac{144}{25} + \frac{81}{25}} = 3$

$S Q_{2} = 1$ .

Mathematics - Parabola Question with Solution | TestHub

Options:(select one or more)

Doubts & Discussion