Question

# Let two tangents are drawn to the curve y2−4(x+y)=3sinθ+4cosθ−15, x,y,θ∈R from the origin whose slopes are m1,m2. If the vertex of the curve is at maximum distance from the origin, then the value of ∣∣∣1m1m2∣∣∣ is

Solution

## Given : y2−4(x+y)=3sinθ+4cosθ−15 ⇒y2−4y+4=4x+3sinθ+4cosθ−11⇒(y−2)2=4(x−11−3sinθ−4cosθ4)⇒(y−2)2=4(x−λ) This is a parabola whose vertex is (λ,2) Now,  λ=11−3sinθ−4cosθ4 We know that,  3sinθ+4cosθ∈[−5,5]⇒λ∈[32,4] Vertex is maximum when λ=4, so the parabola becomes (y−2)2=4(x−4) Now, equation of tangent of slope m is y−2=m(x−4)+1m As the tangents are passing through origin, so putting (0,0) in the equation, we get −2=−4m+1m⇒4m2−2m−1=0 The roots of this equation are m1,m2, then using product of roots, we get m1m2=−14∴∣∣∣1m1m2∣∣∣=4Mathematics

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