Question

A line parallel to the straight line $2 x-y=0$ is tangent to the hyperbola $\left(x^{2} / 4\right)-\left(y^{2} / 2\right)=1$ at the point $\left(x_{1}, y_{1}\right)$. Then $x_{1}{ }^{2}+5 y_{1}{ }^{2}$ is equal to:
a) 6
b) 10
C) 8
d) 5

Answer 1

Hint: Here it is given that the line parallel to the straight line is tangent to the hyperbola. Initially, we have to find the slope and note the changes in coordinates. It is also given that the point lies in the same hyperbola. So we can substitute the values in the general equation of hyperbola to find the required value. Finally equating the two equations will result in the necessary value of the given function

Formula used:
The standard equation of a hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

Complete step by step Solution:
A hyperbola is a collection of points where the separation between each focus is a constant greater than 1. In other words, the locus of a moving point in a plane is such that the ratio of the distance from a fixed point to the distance from a fixed line is always larger than 1.
The slope of the tangent is 2,
Tangent of hyperbola$\dfrac{x^{2}}{4}-\dfrac{y^{2}}{2}=1$ at the point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is
Think about the point $P(x 1, y 1)$ on the given hyperbola. By differentiating the hyperbola equation, it is possible to determine the slope of the tangent at this particular location.
$\dfrac{\mathrm{xx}_{1}}{4}-\dfrac{\mathrm{yy}_{1}}{2}=1 \quad(\mathrm{~T}=0)$
Slope : $\dfrac{1}{2} \dfrac{x_{1}}{y_{1}}=2 \Rightarrow x_{1}=4 y_{1}$……….(1)
A line's slope in mathematics is defined as the ratio of the change in the y coordinate to the change in the x coordinate.

Both the net change in the y-coordinate and the net change in the x-coordinate are denoted by $\Delta \mathrm{x}$and $\Delta y$, respectively.
$\left(x_{1}, y_{1}\right)$ lies on hyperbola
$\Rightarrow \dfrac{x_{1}^{2}}{4}-\dfrac{y_{1}^{2}}{2}=1$ …………………….(2)
From (1) & (2)
$\dfrac{\left(4 y_{1}\right)^{2}}{4}-\dfrac{y_{1}^{2}}{2}=1 \Rightarrow 4 y_{1}^{2}-\dfrac{y_{1}^{2}}{2}=1$
$\Rightarrow 7 y_{1}^{2}=2 \Rightarrow y_{1}^{2}=2 / 7$
Now $x_{1}^{2}+5 y_{1}^{2}=\left(4 y_{1}\right)^{2}+5 y_{1}^{2}$
$=(21) y_{1}^{2}=21 \times \dfrac{2}{7}=6$

Hence, the correct option is (a).

Note: Students made mistakes in understanding the equation of hyperbola.The standard Forms of a Hyperbola with Center (h, k) and the middle rectangle's diagonals and the hyperbola's asymptotes line up. The rectangle's width is 2b and its length is 2a. Each diagonal passes through the centre (h, k), and their slopes are $\pm b a \pm b a$ exists.