Math, asked by tmahajan09876, 9 months ago

If focus of a parabola is (1, 2) and feet of perpendiculars on any two tangents to this parabola from focus are (3, 4) and (4, 6). Then vertex of the parabola is​

Answers

Answered by kanwarpal88109
0

Answer:

I don't know this topic but Mark me AS BRAINLIEST and FOLLOW ME

Answered by poonamsingh1050
0

Answer:

MATHS

avatar

Asked on December 20, 2019 by

Sujatha Punjabi

The locus of the foot of the perpendiculars drawn from the vertex on a variable tangent to the parabola y

2

=4 ax is

Share

Save

ANSWER

The foot of perpendicular from focus to the tangent y

2

=4ax equation of the perpendicular to the tangent is

ty=x+at

2

→ eqn (1)

So, from the focus (a,0) is tx+y=at→(2)

by adding eq (1) & eq (2) we get, fig a

x=0(as(1+t

2

)

=0)

∴ the point of intersection of (1) & (2)

lies on x=0 i.e. on −y axis and this is the tangent at the vertex of the parabola.

As the tangent at the point (P) on a parabola bisects the angle between the focal chord through P and perpendicular from P on the directrix.

∴ the tangent at p(at

2

,2at) is ty=x+at

2

meets x− axis at T(−at

2

,0)

Hence, from the figure we can analyse that

ST=SA+AT=a(1+t

2

)

also SP=

a

2

(i+t

2

)+4a

2

t

2

=a(1+t

2

)=ST

which means ∠MPT=∠PTS=∠SPT=TP

Let p(at

2

,2a) be a point on the parabola y

2

=4ax

equation of the tangent P is ty=x+at

2

Point of intersection of the tangent with the directrix is ratio where (−a,(at−a)/t)

Now, slope of SP is (2at−0)/(at

2

−a)=2t/(t

2

−1)

parallel slope of SK is (at−at/0)/(−a−a)=−(t

2

−1)/2t

⇒ (Slope of the SP) (Slope of SK)=−1 there by SP is perpendicular to SK i.e., ∠KSP=90

o

as we know that Tangent at teh extremities of any focal chord intersect at right angels on the directrix.

is p(at

2

,2at),p(at

1

2

,2at

1

) are end points we get t.t

1

=−1⇔ eq of tangent ty=x+at

2

& t

1

y=x+at

1

2

if these tangent intersect ar a point (h,k) then

h=att

1

and k=a(t+t

1

).

∴ the tangent are perpendicular , tt

1

=−1⇒h−a

∴ the locus of the point (h,k) is x=−a and it is the equation of directrix

Similar questions