IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

neeraj_agarwal_1990

The tangent and normal at the point P(at²,2at) to the parabola y²=4ax meet the x-axis in T and G respectively,then angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through P,T,G is?

[ans:tan^-1(t)]
I'm getting different ans

ankursingh

abhishekpunpale

according to your question a tangent cuts x-axis in 2 different points

is at possible??

avdesh100

neeraj i suppose,
u can find the equation of the circle by the given three points and then find the tangent due to P on the circle and then find the angle b/w them by the formula of tanQ!

neeraj_agarwal_1990

avdesh...I tried it but am getting different ans..

avdesh100

then ur answer is correct...anyways i think this is lengthy but sure method

miyo

Hi neeraj,

The circle is nothing but the 1 with diameter as TG and center at S(a,0) ;

Let the

PTS =

;{ where Tan

= 1 / t }

Let the angle bt tangents be

;

So, ext angle

PSG = {

+(

/2 -

)}

Just expand..Tan

PSG { = 2t / (t²-1) } ;

Solve for cot

;{u get it as 1/t }

So,

= Tan^-1(1/t) ....ans.

iitkgp_bipin

You just draw the figure.Since tangent and normal at P are perpendicular to each other,so

TPG = 90⁰.
This is the angle contained in a semicircle and hence TG would be the diameter of the circle passing through P,T and G.

In equation of tangent and normal put y=0 to get T and G.
Then T is (-at²,0) and G is (2a - at²,0)
Hence centre of the circle is (a,0) which is the focus of parabola.

Slope of normal to the circle at P = 2at/(at²-a) = 2t/(t²-1)
Hence slope of tangent to the circle at P = -1/(slope of normal) = (1-t²)/2t

Also slope of tangent at P to parabola = 1/t

Now slopes of two lines are known and you can easily find the angle between them.

Best Wishes

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