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Area between Two Curves

If the lines ...

Question

If the lines $(y - b) = m_{1} (x + a)$ and $(y - b) = m_{2} (x + a)$ are the tangents of $y^{2} = 4 a x$ , then:

m_{1} + m_{2} = 0

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m_{1} . m_{2} = 1

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m_{1} m_{2} = - 1

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m_{1} + m_{2} = 1

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Solution

The correct option is A $m_{1} m_{2} = - 1$
we know that,

$y = m x + \frac{a}{m}$ represents equation of the tangent to parabola

$y^{2} = 4 a x$

if it passes through a point $(x_{1}, y_{1})$

we have
$y_{1} = m x_{1} + \frac{a}{m}$ OR $m y_{1} = m^{2} x_{1} + a$

$\Rightarrow m^{2} x_{1} - m y_{1} + a = 0$ ..........(1) which gives 2 values of m in general.

In the present case equations of the tangents to parabola
$y^{2} = 4 a x$ are given as

$y - b = m_{1} (x + a)$ and $y - b = m_{2} (x + a)$

$\Rightarrow$ tangents are passing through $(- a, b)$ and have slopes $m_{1}$ and $m_{2}$

In equation (1) we can substitute $x_{1} = - a$ and $y_{1} = b$

we have
$m^{2} (- a) - m (b) + a = 0$ which will have roots as $m_{1}$ and $m_{2}$

the above equation can be written as

$a m^{2} + b m - a = 0$

product of roots

$m_{1} \times m_{2} = \frac{- a}{a} = - 1$

$\Rightarrow$ one slope is negative reciprocal of other

$\Rightarrow$ tangents are perpendicular

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If the lines (y−b)=m1(x+a) and (y−b)=m2(x+a) are the tangents of y2=4ax, then:

If the lines $(y - b) = m_{1} (x + a)$ and $(y - b) = m_{2} (x + a)$ are the tangents of $y^{2} = 4 a x$ , then: