Class 11 Conic Section Exercise 10.2 Solution

In each of the following Exercises 16 to 22, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

16. y² = 12x

$?$ ae = 2b

$\frac{4b^2}{a^2}=e^2$  

Or 4 (1 – e²) = e²

4 = 5e² -> $e=\frac{2}{\sqrt[]{5}}$

If two circles intersect at two distinct points

->|r₁ – r₂| < C₁C₂ < r₁ + r₂

| r – 2| <  $\sqrt[]{9+16}$ < r + 2

|r – 2| < 5                     and r + 2 > 5

–5 < r – 2 < 5                    r > 3                      … (2)

–3 < r < 7                                                        … (1)

From (1) and (2)

3 < r < 7

x² – y² cosec²q = 5 $\frac{x^2}{1}-\frac{y^2}{si{n}^2\theta }=5$                        

x² cosec²q + y² = 5  $\frac{x^2}{si{n}^2\theta }+\frac{y^2}{1}=5$        

$e_H=\sqrt[]{7}{e}_e$                  

and $e_H=\sqrt[]{1+\frac{si{n}^2\theta }{1}}$ &n

Slope of axis = $\frac{1}{2}$

$y-3=\frac{1}{2}\left(x-2\right)$              

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

            α + 1.6 = 4 ⇒ α = 2.4

            β + 2.8 = 6 ⇒

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

K = $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e2 – 1)

$e=\sqrt[]{4}=2$