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  • Tangents drawn from the point to the parabola <img alt="\mathrm{y^2=8 x}" src="https://entrancecorner.oncodec

Tangents drawn from the point (-8,0) to the parabola \mathrm{y^2=8 x} touch the parabola at \mathrm{P} and \mathrm{Q}. If \mathrm{F} is the focus of the parabola. then the area of the triangle PFQ(in sq. units) is equal to

Option: 1

48


Option: 2

35


Option: 3

30


Option: 4

41


Answers (1)

best_answer

Let \mathrm{\left(2 t^2, 4 t\right)} be the point on the parabola and the focus is \mathrm{(2,0)}
Now, slope of the tangent to the parabola \mathrm{y^2=8 x} is given by
\mathrm{ 2 y \cdot \frac{d y}{d x}=8 \Rightarrow \frac{d y}{d x}=\frac{4}{y} }

\mathrm{ \left.\frac{d y}{d x}\right|_{\substack{x=2 t^2 \\ y=4 t}}=\frac{4}{4 t}=\frac{1}{t} }         ......(i)
But the slope of the tangent passing through \mathrm{ \left(2 t^2, 4 t\right) and (-8,0) } is given by

\mathrm{ m =\frac{y_2-y_1}{x_2-x_1} \Rightarrow m=\frac{0-4 t}{-8-2 t^2} }

\mathrm{ \Rightarrow m =\frac{4 t}{8+2 t^2} }      (ii)

Equating (i) and (ii), we have

\mathrm{ \frac{1}{t}=\frac{4 t}{8+2 t^2} \Rightarrow \frac{1}{t}=\frac{2 t}{4+t^2} }

\mathrm{ \Rightarrow 4+t^2=2 t^2 \Rightarrow 4=t^2 \Rightarrow t= \pm 2 }

\mathrm{ For \: \: t=2, point \: is (8,8) and\: for \: t=-2, point\: is (8,-8) }

\mathrm{ \text{So, the points are} \: P(8,8), Q(8,-8), F(2,0) }
\mathrm{ \therefore \quad \text { Area of } \triangle P F Q=\frac{1}{2}|8(-8)+8(-8)+2(8+8)| }

\mathrm{ \quad=\frac{1}{2}|-128+32|=\frac{1}{2}|-96|=48 \text { sq. units }}








 

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