Equation of tangent to #160; y^2=4ax is #160; y=mx+a/m Let it passes through P(h, k) ∴ m^2h mk+a=0 (ii) #160; tanθ _0=m_1+m_2/1 m_1m_2=k/ ...

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Standard XII

Mathematics

Integration of Trigonometric Functions

ii θ 1+θ 2=...

Question

(ii) $θ_{1} + θ_{2} = θ_{0}$ (const)

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Solution

Equation of tangent to $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$
Let it passes through P(h, k)
$∴ m^{2} h - m k + a = 0$
(ii) $tan θ_{0} = \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} = \frac{k / h}{1 - a / h}$ $\Rightarrow y = (x - a) tan θ_{0}$

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(ii) θ1+θ2=θ0 (const)

Equation of tangent to y2=4ax is y=mx+amLet it passes through P(h, k)∴m2h−mk+a=0(ii) tanθ0=m1+m21−m1m2=k/h1−a/h ⇒y=(x−a)tanθ0

(ii) $θ_{1} + θ_{2} = θ_{0}$ (const)

Equation of tangent to $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$
Let it passes through P(h, k)
$∴ m^{2} h - m k + a = 0$
(ii) $tan θ_{0} = \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} = \frac{k / h}{1 - a / h}$ $\Rightarrow y = (x - a) tan θ_{0}$