If the line alpha x+2y=1,where alpha in mathb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the line is $y = \frac{1-\alpha x}{2}$.Substituting this into the hyperbola equation $x^2 - 9y^2 = 9$:$x^2 - 9\left(\frac{1-\alpha

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(B) The equation of the line is $y = \frac{1-\alpha x}{2}$.Substituting this into the hyperbola equation $x^2 - 9y^2 = 9$:$x^2 - 9\left(\frac{1-\alpha x}{2}\right)^2 = 9$$x^2 - \frac{9}{4}(1 - 2\alpha x + \alpha^2 x^2) = 9$$4x^2 - 9 + 18\alpha x - 9\alpha^2 x^2 = 36$$(4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$Since the line does not intersect the hyperbola,the discriminant $D $D = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) $324\alpha^2 + 180(4 - 9\alpha^2) $324\alpha^2 + 720 - 1620\alpha^2 $-1296\alpha^2 + 720 $1296\alpha^2 > 720$$\alpha^2 > \frac{720}{1296} = \frac{5}{9}$$|\alpha| > \frac{\sqrt{5}}{3} \approx 0.745$Comparing with the options,the only value greater than $0.745$ is $0.8$.

If the line $\alpha x+2y=1$,where $\alpha \in \mathbb{R}$,does not meet the hyperbola $x^{2}-9y^{2}=9$,then a possible value of $\alpha$ is:

Similar Questions

Find the equation of the hyperbola whose foci are $(6, 5)$ and $(-4, 5)$ and eccentricity is $5/4$.

For the hyperbola $x^2-y^2-4x+2y+c=0$,if the focus is $S(2+2\sqrt{2}, k)$ and the directrix that is adjacent to $S$ is $x=2+\sqrt{2}$,then $c=$

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16x^{2}-9y^{2}+32x+36y-164=0$ and its foci is:

If the tangent drawn at a point $P(t)$ on the hyperbola $x^2-y^2=c^2$ cuts the $X$-axis at $T$ and the normal drawn at the same point $P$ cuts the $Y$-axis at $N$,then the equation of the locus of the midpoint of $TN$ is

The equation of the tangent to the conic $x^2 - y^2 - 8x + 2y + 11 = 0$ at the point $(2, 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module