EllipseeasyFree

Minimum of (r+5-4cos)^2+(r-3sin)^2 Expression | JEE

JEE Maths question with a full step-by-step solution.

Question
The minimum value of {(r+5−4∣cos⁡θ∣)2+(r−3∣sin⁡θ∣)2}\left\{(r+5-4|\cos\theta|)^2+(r-3|\sin\theta|)^2\right\}, ∀ r,θ∈R\forall\,r,\theta\in\mathbb{R}, is:
A00correct
B22
C33
Dnone of these
Solution
Step 1: Read the expression as a squared distance. The point (r+5, r)(r+5,\ r) lies on the line y=x−5y=x-5, and the point (4∣cos⁡θ∣, 3∣sin⁡θ∣)(4|\cos\theta|,\ 3|\sin\theta|) lies on the ellipse x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1. The bracket is the squared distance between these two points. Step 2: So the minimum equals the squared shortest distance between the line y=x−5y=x-5 and the ellipse x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1. Step 3: The tangents to the ellipse with slope 11 are y=x±16⋅1+9=x±5y=x\pm\sqrt{16\cdot1+9}=x\pm5. Thus y=x−5y=x-5 is itself a tangent to the ellipse, so the shortest distance is 00. ∴\therefore the minimum value is 00. Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Ellipse · easy
The equation x210−a+y24−a=1\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1 represents an ellipse if:
Ellipse · easy
The eccentricity of the ellipse (x−3)2+(y−4)2=y29(x-3)^2+(y-4)^2=\dfrac{y^2}{9} is:
Ellipse · medium
An ellipse having foci at (3,3)(3,3) and (−4,4)(-4,4) and passing through the origin has eccentricity equal to:
Ellipse · medium
The foci of an ellipse are (−2,4)(-2,4) and (2,1)(2,1). The point (1,236)\left(1,\dfrac{23}{6}\right) is an extremity of the minor axis. What is the value of the eccentricity?

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath