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# Two tangents are drawn from a point (-1,2) to a parabola ${y^2} = 4x$ Find the angle between the tangents.

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Hint: Compare the given equation of parabola with the general equation, you’ll get the value of ‘a’. Now, find out the value ${m_1}$ and ${m_2}$ and find out the angle.

The general equation of parabola is ${y^2} = 4ax$

Equation of tangent to this parabola is $y = mx + \dfrac{a}{m}$

Given equation of parabola is ${y^2} = 4x$ and clearly a = 1

So the equation of tangent will become $y = mx + \dfrac{1}{m}$

Now two tangents are drawn from the point ( - 1,2) so in place of x and y substitute the values,

So equation of tangent is $2 = - m + \dfrac{1}{m}$

On solving this we get a quadratic equation ${m^2} + 2m - 1 = 0$

Using $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Roots are coming out as $\dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{2}$

So from this we are getting two slopes that is ${m_1} = - 1 + \sqrt 2$ and ${m_2} = - 1 - \sqrt 2$

Now $tan\theta= \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$

Using above $tan\theta$ =$\left| {\dfrac{{ - 1 + \sqrt 2 + 1 + \sqrt 2 }}{{1 + \left( { - 1 + \sqrt 2 } \right)\left( { - 1 - \sqrt 2 } \right)}}} \right|$

Now it simplifies to $\left| {\dfrac{{2\sqrt 2 }}{{1 + \left( { - {1^2} - {{\sqrt 2 }^2}} \right)}}} \right|$ using (a - b)(a + b) =${a^2} - {b^2}$

On solving this we are getting a zero on denominator hence our $tan\theta= \infty$ hence $\theta= \dfrac{\pi }{2}$

Note -Always remember the equation of tangent with respect to the general equation of parabola and just satisfy the points to this equation of tangent.

Last updated date: 23rd Sep 2023
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