Parabola (S .N. Dey ) | Ex-4

In the previous article , we have solved 6 Long Answer Type Questions of Parabola Chapter. In this article, we have solved 6 more Long answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.

$7.~$ Show that the product of the ordinates of the ends of a focal chord of the parabola $~y^2 = 4ax~$ is constant.

Solution.

We know that the co-ordinates of extremities of the focal chord are $~(at^2,2at)~$ and $~(a/t^2,-2a/t).$

So, the product of the ordinates

$=2at \times \left (\frac{-2a}{t} \right)=-4a^2=\text{constant.}$

$8.~$ If a straight line passing through the focus of the parabola $~y^2=4ax~$ intersects the parabola at the points $~(x_1,y_1)~$ and $~(x_2, y_2)~$ then prove that, $~y_1y_2+4x_1x_2=0.$

Solution.

Let $~(x_1,y_1) \equiv(at_1^2,2at_1),~~(x_2,y_2)=(at_2^2,2at_2).$

Since the chord of the given parabola passes through the focus $~(a,0)~$ i.e., the chord is the focal chord.

So, $~t_1t_2=-1.$

$\therefore~~ y_1y_2+4x_1x_2\\=(2at_1)(2at_2)+4 \cdot (at_1^2)(at_2^2)\\=4a^2 \cdot t_1t_2+ 4a^2 \cdot t_1^2t_2^2 \\=4a^2(-1)+4a^2(-1)^2\\=-4a^2+4a^2\\=0.$

$9.~$ Find the equation of the circle passing through the origin and the foci of the parabolas $~y^2 = 8x~$ and $~x^2 = 24y.$

Solution.

The given equations of parabolas are

$~y^2=8x=4 \times 2 \times x \rightarrow(1), \\ x^2=24y =4 \times 6 \times y \rightarrow(2).$

The focus of the parabola $~(1)~$ is $~(2,0)~$ and that of $~(2)~$ is $~(0,6).$

Let the equation of the circle be

$~x^2+y^2+2gx+2fy+c=0 \rightarrow(3)$

Since the circle $~(3)~$ passes through the points $~(0,0),~(2,0),~(0,6)~$ , so

$~0+0+0+0+c=0 \Rightarrow c=0 ,\\~ 2^2+0+2g \times 2+0+0=0 \Rightarrow g=-\frac 44=-1, \\~ 0+6^2+0+2f \times 6+0=0 \Rightarrow f=-\frac{36}{12}=-3.$

Hence, the equation of the required circle is

$x^2+y^2+2 \times (-1) x+2 \times (-3)y=0 \\ \text{or,}~~x^2+y^2-2x-6y=0.$

$10.~$ Find the equation of the circle on $~SC~$ as diameter, where $~S~$ is the focus of $y^2 = 12x~$ and $~C~$ is the centre of $~x^2+ y^2-18x-16y+45= 0.~$ Also find the length of the chord of the circle lying along the $x-$ axis.

Solution.

The co-ordinates of the focus $~(S)~$ of the parabola $~y^2=12x=4 \times 3 \times x~$ are $~(3,0).~$

The centre $~(C)~$ of the given circle $~x^2+y^2-18x-16y+45=0~$ is $~(9,8).$

Now, the equation of the circle having diameter with extremities $~S(3,0)~$ and $~C(9,8)~$ is

$~(x-3)(x-9)+(y-0)(y-8)=0 \\ \text{or,}~~ x^2+y^2-12x-8y+27=0.$

So, the length of the chord of the circle lying along the $~x-$ axis is

$=2\sqrt{6^2-27}=2\times \sqrt{9}=6~~\text{unit}.$

$11.~~ Q~$ is any point on the parabola $~y^2 = 4ax;~ QN~$ is the ordinate of $~Q~$ and $~P~$ is the mid-point of $~QN.~$ Prove that the locus of $~P~$ is a parabola whose latus rectum is one-fourth that of the given parabola.