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Parabola
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Parabola
E-3 IF Three distinct and real normals can be drawn to y2=8x from the point (a,0) then show that a>4.

ans:- Equation of normal in terms of m is y=mx-4m-2m3=0

it passes through (a,0) then am-4m-2m3=0

m(a-4-2m2)=0

= m=0, m2=

for three distinct normal , (a-4)>0

=a>4



E-4 if y+b=m,(x+a) and y+b=m(x+a) are two tangents to the parabola y24ax, then show that m1m2=-1

which lies on the directrix x+a=0. hence th etwo tangents intersect on directrix which we know is the locus of perpendicular tangents .

hence m1.m2=-1

E-5 show that the parametric representation (2+t2,2t+1)

represents a parabola with vertex at (2,1).



Ans:- x=2+t2, y=2t+1.

Eliminating t= we get (y-1) 2 =4(x-2)

i.e, a parabola with vertex at (2,1).



Ans: Equation of the given parabola can be written as,

9x2+12x+4+18y-18=0

i.e, (3x+2)2=18(y-1)



Equation of tangent to the above parabola can be written as-



IF the tangent passes through (0,1) then we have,

0.m2 -4m-3=0

gives,m=-

hence equation of the required lines are,



i.e, 12x+9y-1=0

and y-1=0



E-8: the tangents to the parabola y2 =4ax at p(at2,2at) and q(at22,2at2) intersect at r. prove that the area of the triangle

PQR is



Ans:







equation of tangents at p(at2,2at1) and Q(at22,2at2)

are, t1y=x+at12............(1)

and t2y=x+at22

since point of intersection of (1) and (2) is r(at1t2) (at1t2,a(t1+t2)



E-6 the normals with slopes m1, m2, and m3 are drawn from apoint p not on the axis of the parabola y2

=4ax ifv m1 and m2= results in the locus of p being a part of parabola, find the value of

Ans:







Any normal of the parabola y2=4x with slope m is

y=mz-2m-m3

thus, m1.m2.m3=-k

m3=-k(m1m2=

=

m3 is aroot of(1) then,



=k3+(2-h)k2-k3=0

locus of p(h,k) is,

y3+(2-x)y2-y3=0

(p does not lie on the axis of the parabola)

=y2=2x-22+3

it is a part of the parabola y2=4x

then 2=4

and -22+3=0

=-2=0

=2



E-7: find the equation of aline which touches the parabola

9x2+12x+18y-14=0

and passes through the point (0,1).

AREA OF TRAINGLE: .



Expanding with respect to first row-



.



E-9: Find the length of the normal chord to the parabola y2=4x which substends a right angle

at the vertex Ans:- a=1 for parabola PQ being normal chord.





PQ substend a right angle at vertex,





E - 10. An equilateral triangle SAB is inscribed in the parabola y2 = 4ax Having its focus at 'S'.

It chord AB lies towards the left of S. Then find the side length of this triangle.

Ans:





let A=(at12,2at1),B=(at22-2at1)

we have





Q– Prove that 9x2 - 24xy + 16y2 - 20x - 15 y -

60 = 0 represents a parabola. Also find its focus and directrix.



Ans: Here h2-ab=(-12)2-9x16=144-144=0 also

the equation represents a parabola Now, the equation is(3x-4y)2=5(4x+3y+12) clearly

, the lines 3x-4y=0 and 4x+3y+12=0 are perpendicular to each other.so, let

the equation of the parabola becomes -



Now if

then we havefrom the equations of tranformation in(1)



The equation of directrix is,





The directrix is
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