Conic Sections Question 146
Question: The locus of a point P( $ \alpha $ , $ \beta $ )moving under the condition that the line $ y=\alpha x+\beta $ is a tangent to the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ is
Options:
A) An ellipse
B) A circle
C) A parabola
D) A hyperbola
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The tangent to the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ is $ y=mx\pm \sqrt{a^{2}m^{2}-b^{2}} $ Given that $ y=\alpha x+\beta $ is a tangent of the hyperbola. So $ m=\alpha $ and $ a^{2}m^{2}-b^{2}={{\beta }^{2}} $
$ \therefore a^{2}{{\alpha }^{2}}-b^{2}={{\beta }^{2}} $ The locus is $ a^{2}x^{2}-y^{2}={b^{2,}} $ which is a hyperbola.
BETA
