Correct Answer – Option 3 : 3/5

Concept:

Ellipse:

Calculation:

Given: \(\rm \frac{x^2}{16} + \frac{y^2}{25} = 1\)

Compare with the standard equation \(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

So, a² = 16 and b² = 25 ⇔ a = 4 and b = 5 (b > a)

So, eccentricity = \(\rm \sqrt{1-\frac{a^2}{b^2}}\)

= \(\sqrt{1-\frac{16}{25}}\)

= 3/5

Let P(x₁, y₁) be a point on the line 2x – 3y + 4 = 0. Therefore

2x₁ – 3y₁ + 4 = 0 ….(1)

Now, the chord of contact of (x₁, y₁) with respect to y² + 4ax is

yy₁ – 2a(x + x₁) = 0 ….(2)

From Eqs. (1) and (2), we get

yy₁ – 2ax – a(3y₁ – 4) = 0

y₁(y – 3a) 2a(x – 2) = 0  …(3)

Eq. (3) represents the lines passing through the fixed point which is the intersection of the lines x = 2 and y = 3a. Hence, the fixed point is (2, 3a).

Correct Answer – Option 2 : \(\rm {(x-1)^2\over 4}-{y^2\over 12} = 1\)

Concept:

The standard equation of a hyperbola:

\(\rm {(x-h)^2\over a^2}-{(y-k)^2\over b^2} = 1\)

where 2a and 2b are the length of the transverse axis and conjugate axis respectively and centre (h, k)

Note: The centre is the midpoint of the 2 foci.

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

Length of latus recta = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2+b^2}\)

Calculation:

Given foci are (5,0) and (-3,0)

Center = \(\rm \left({5+(-3)\over2},{0+0\over2}\right)\) = (1, 0)

Now distance of focus from the center = \(\rm \sqrt{a^2+b^2}\)

⇒ \(\rm \sqrt{(5-1)^2+(0-0)^2} = \rm \sqrt{a^2+b^2}\)

⇒ a² + b² = 16               …(i)

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

⇒ 2 = \(\rm \sqrt{16}\over a\) 

⇒ a = 2

Putting it in (i)

⇒ 4 + b² = 16

⇒ b² = 12

a² = 4

The equation of the hyperbola

\(\rm {(x-h)^2\over a^2}-{(y-k)^2\over b^2} = 1\)

⇒ \(\rm {(x-1)^2\over 4}-{(y-0)^2\over 12} = 1\)

⇒ \(\boldsymbol{\rm {(x-1)^2\over 4}-{y^2\over 12} = 1}\)

Let P(h, k) be a point in the plane of y² = 4ax. It is known that the normal to parabola at (at², 2at) is tx + y = 2at + at³. This normal passes through P(h,k

th + k = 2at + 2at + at³ 

at³ + (2a – h)t – k = 0   …..(1)

Equation (1) is a cubic equation in t and hence, in general, it has three roots and hence, there are three points on the curve at which normals drawn to the parabola are concurrent at P(h, k). 

Correct Answer – Option 1 : \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{16}}} = 1\)

CONCEPT:

The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\) are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

CALCULATION:

Here, we have to find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8.

By comparing the foci (± 5, 0) with (± ae, 0)

⇒ ae = 5

∵ Length of the conjugate axis is given by 2b

⇒ 2b = 8

⇒ b = 4

As we know that, eccentricity of a hyperbola is given by \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)

⇒ a²e² = a² + b²

⇒ 25 = a² + 16

⇒ a² = 9

So, the equation of the required hyperbola is \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{16}}} = 1\)

Hence, option A is the correct answer.