Test Index

Parabola Part 1

Section: Mathematics

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Question : 54 of 100

Marks: +1, -0

Let

L_1

be a tangent to the parabola

y^{2}=4(x+1)

L_2

be a tangent to the parabola

y^{2}=8(x+2)

L_1

L_2

intersect at right angles. Then

L_1

L_2

meet on the straight line:

[6 Sep 2020 Shift 1]

$x+3=0$
$x+2y=0$
$2x+1=0$
$x+2=0$

Solution:

y^{2}=4(x+1)

equation of tangent

y=m(x+1)+\frac{1}{m}

y=mx+m+\frac{1}{m}

y^{2}=8(x+2)

equation of tangent

y=m'(x+2)+2m'

y=m'x+2\left(m+\frac{1}{m'}\right)

since lines intersect at right angles

\therefore mm'=-1

Now

y=mx+m+\frac{1}{m}

...(1)

y=m'x+2\left(m'+\frac{1}{m}\right)

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

...(2)

From equation (1) and (2)

mx+m+\frac{1}{m}=-\frac{1}{m}x-2\left(m+\frac{1}{m}\right)

\left(m+\frac{1}{m}\right)x+3\left(m+\frac{1}{m}\right)=0

\therefore x+3=0

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