lf y = mx+ c be a tangent to hyperbola x2λ2−y2λ3+λ2+λ2=1 then least value of 16m2 equal to

c2=a2m2−b2⇒c2=λ2m2−λ3+λ2+λ2c2≥0⇒m2≥λ2+λ+12λ2+λ+1 has minimum value 34⇒m2≥916

lf y = mx+ c be a tangent to hyperbola x2λ2−y2λ3+λ2+λ2=1 then least value of 16m2 equal to

c2=a2m2−b2⇒c2=λ2m2−λ3+λ2+λ2c2≥0⇒m2≥λ2+λ+12λ2+λ+1 has minimum value 34⇒m2≥916

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

lf y = mx+ c be a tangent to hyperbola $\frac{x^{2}}{λ^{2}} - \frac{y^{2}}{{(λ^{3} + λ^{2} + λ)}^{2}} = 1$ then least value of 16m² equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} c^{2} = a^{2} m^{2} - b^{2} \Rightarrow c^{2} = λ^{2} m^{2} - {(λ^{3} + λ^{2} + λ)}^{2} \\ c^{2} \geq 0 \Rightarrow m^{2} \geq {(λ^{2} + λ + 1)}^{2} \\ λ^{2} + λ + 1 has minimum value \frac{3}{4} \Rightarrow m^{2} \geq \frac{9}{16} \end{array}$