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Latus Rectum

If y+b = m1...

Question

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

m_{1} m_{2} = - 1

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m_{1} m_{2} = 1

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m_{1} + m_{2} = 0

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None of these

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Solution

The correct option is A $m_{1} m_{2} = - 1$
Solution:- (A) $m_{1} . m_{2} = - 1$
Given:-
$y^{2} = 4 a x$
By checking $x + a = 0 \Rightarrow x = - a$
This means that the two tangents passes through the directrix and hence are perpendicular to each other.
Since they are perpendicular, the product of their slopes will be $- 1$ , i.e.,
$m_{1} . m_{2} = - 1$

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Similar questions

y + b = m_{1} (x + a), y + b = m_{2} (x + a)

y^{2} = 4 a x,

m_{1}

m_{2}

y^{2} = 4 a x

y + 3 = m_{1} (x + 2)

y + 3 = m_{2} (x + 2)

y^{2} = 8 x

m_{1} m_{2} = - 1

y = m_{1} x + c

y = m_{2} x + c

y^{2} + 4 a (x + a) = 0

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

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

y^{2} = 4 a x

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