17 Questions around this concept.
The equation of tangent to the ellipse $\frac{x^2}{50}+\frac{y^2}{32}=1$ which passes through a point $(15,-4)$ is
The values of $\lambda$ for which the line $y=2 x+\lambda_{\text {is a chord to hyperbola }} x^2-y^2-2 x+2 y-1=0$ is/are
Hyperbola : $\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Line: $\quad y=m x+c$
After solving Eq. (i) and Eq. (ii)
$
\begin{aligned}
& \quad \frac{x^2}{a^2}-\frac{(m x+c)^2}{b^2}=1 \\
& \Rightarrow \quad\left(a^2 m^2-b^2\right) x^2+2 \mathrm{mca}^2 x+c^2 a^2+a^2 b^2=0
\end{aligned}
$
Above equation is quadratic in $x$
The line will cut the hyperbola in two points may be real, coincident or imaginary, that depends on the value of Discriminant, D.
1. If $D>0$, then two real and distinct roots which means two real and distinct points of intersection of the line and the hyperbola. In this case, the line is secant (chord) to the hyperbola.
2. If $\mathrm{D}=0$, then equal real roots which means the line is tangent to the hyperbola. Solving $\mathrm{D}=0$ we get the condition for tangency, which is $c^2=a^2 m^2-b^2$
3. If $D<0$, then no real root which means the line and hyperbola do not intersect.

"Stay in the loop. Receive exam news, study resources, and expert advice!"
Coordinate Geometry (Arihant)
Page No. : 565
Line : 9
151+ Downloads
752+ Downloads
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Last Date to Apply: 29th April | Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements
100% Placement Assistance | Avail Merit Scholarships | Highest CTC 43 LPA
Last Date to Apply: 26th April | NAAC A++ Accredited | NIRF Rank #3
Application Deadline: 18th April | Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships
World-class and highly qualified engineering faculty. High-quality global education at an affordable cost
Explore on Careers360
Student Community: Where Questions Find Answers