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Question: Equation of a tangent to the hyperbola \[5{{x}^{2}}-{{y}^{2}}=5\] and which passes through an extern...

Equation of a tangent to the hyperbola 5x2−y2=55{{x}^{2}}-{{y}^{2}}=5 and which passes through an external point (2, 8) is
A. 3x-y+2=0
B. 3x+y-14=0
C. 23x-3y-22=0
D. 5x-4y+22=0

Explanation

Solution

Hint:Differentiate the equation of hyperbola and find the slope of tangent of hyperbola, where m=dydxm=\dfrac{dy}{dx}. Then find the equation of tangent by using slope. Substitute the point (2, 8) in the equation of tangent to get the required equation.

Complete step-by-step answer:
Given us the equation of the hyperbola⇒5x2−y2=5−(1)\Rightarrow 5{{x}^{2}}-{{y}^{2}}=5-(1)
The slope of tangent of the hyperbola is denoted by m.
∴m=dydx\therefore m=\dfrac{dy}{dx}, which is equal to the differential.
Now differentiate equation (1) with respect to ‘x’.
5x2−y2=55{{x}^{2}}-{{y}^{2}}=5

& \Rightarrow 10x-2y.\dfrac{dy}{dx}=0 \\\ & \therefore +2y\dfrac{dy}{dx}=10x \\\ & \therefore \dfrac{dy}{dx}=\dfrac{10x}{2y} \\\ & \dfrac{dy}{dx}=\dfrac{5x}{y} \\\ \end{aligned}$$ $$\therefore $$The slope of tangent of the hyperbola, $$m=\dfrac{dy}{dx}$$ $$m={{\left( \dfrac{dy}{dx} \right)}_{\left( 2,8 \right)}}$$as the equation of tangent to the hyperbola passes through the point (2, 8). $$\begin{aligned} & \therefore m={{\left( \dfrac{dy}{dx} \right)}_{\left( 2,8 \right)}} \\\ & m={{\left( \dfrac{5x}{y} \right)}_{\left( 2,8 \right)}}=\dfrac{5\times 2}{8}=\dfrac{10}{8}=\dfrac{5}{4} \\\ & \therefore m=\dfrac{5}{4} \\\ \end{aligned}$$ Equation of tangent is given by the formula. $$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$$ $$\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,8 \right)$$and $$m=\dfrac{5}{4}$$ $$\Rightarrow \left( y-8 \right)=\dfrac{5}{4}\left( x-2 \right)$$ Cross multiplying and simplify it, $$4\left( y-8 \right)=5\left( x-2 \right)$$ Open the brackets, $$\begin{aligned} & 4y-32=5x-10 \\\ & \Rightarrow 5x-4y+32-10=0 \\\ & 5x-4y+22=0 \\\ \end{aligned}$$ Hence, the correct option is (D). Note: The slope of tangent is given by $$m=\dfrac{dy}{dx}$$ by differentiating the equation of parabola. So, it's important to find the derivative of the equation of hyperbola.Students should remember equation of tangent given by $$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$$