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Hyperbola — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsHyperbola5 Marks
Question
focus is $(2, 2),$ directive is $x + y = 9$ and eccentricity $= 2.$
✓

Answer

Let $S (2, 2)$ be the focus and $P(x, y)$ be a point on the hyperbola.
Draw PM perpendicular from $P$ on the directrix. Then, by definition
$sP = ePM$
$\Rightarrow sP^2 = e^2PM^2$
$\Rightarrow(\text{x}-2)^2+(\text{y}-2)^2=2^2\Big[\frac{\text{x}+\text{y}-9}{\sqrt{1^2+1^2}}\Big]$ $\Big[\because\ \text{e}=\frac{4}{3}\Big]$}
$\Rightarrow\ \text{x}^2+4-4\text{x}+\text{y}^2+4-4\text{y}=\frac{4[\text{x+y}-9]^2}{2}$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = 2[x + y - 9]^2$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = 2[x + y + (-9) + 2 \times x \times y + 2 \times y \times (-9) + 2 \times (-9) \times x]$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = 2[x^2 + y^2 + 81 + 2xy - 18y + 18x]$
$\Rightarrow x^2 + y^2 - 4x - 4y + 8 = [2x^2 + 2y^2 + 162 + 4xy - 36y + 36x]$
$\Rightarrow 2x^2 - x^2 + 2y^2 - y^2 + 4xy - 36x + 4x - 36y + 4y + 162 - 8 = 0$
$\Rightarrow x^2 + y^2 - 4xy - 32x - 32y + 154 = 0$
This is the required equation of the hyperbola.

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