Conic Sections - 2 - Free MCQ Practice Test with solutions, JEE Maths

For the ellipse

So the foci are at 

c = 0; therefore Option B is correct.

The derivative of y = (7/2) x² is y' = 2·(7/2) x = 7x.

A horizontal line has slope 0; for tangency the slope at the point of contact must be 0, so 7x = 0 ⇒ x = 0.

The value of y at x = 0 on the parabola is y = (7/2)·0² = 0, hence the horizontal tangent has c = 0.

Thus the correct answer is B (0).

Given the equation is,
2x²+3y²−8x−18y+35=K
Or, 2{x²−4x+4} + 3{y²−6y+9}=K
Or, 2(x−2)² + 3(y−3)² =K.
From the above equation it is clear that if K>0 then the given equation will represent an ellipse and for K<0, no geometrical interpretation.
Also if K=0 then the given equation will be reduced to a point and the point will be (2,3).

P(x,y) = [(e^t + e^-t)/2 , (e^t - e^-t)/2]
(e^t + e^-t)/2 = x --------------------------(1)
(e^t - e^-t)/2 = y --------------------------(2)
Adding (1) & (2)
2e^t = 2x + 2y
e^t = x + y
Eq (1) e^t + e^-t = 2x
e^t + 1/e^t = 2x
(e^t)² + 1 = 2x*e^t
(x+y)² + 1 = 2x(x+y)
x² + y² + 2xy + 1 = 2x² + 2xy
x² + y² + 1 = 2x²
(x²)/(1)² - (y²)/(1)² = 1  {which represents hyperbola equation} 

Distance of 'c' units from (2,3)
Let the no: of points be (x,0)
By distance formula
{(2−x)²+(3−0)²}=c
 4−4x+x²+9=c
⇒x²−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.

Divide both sides of 9x² - 16y² = 144 by 144 to get x²/16 - y²/9 = 1.

This is a hyperbola of the form x²/a² - y²/b² = 1, so a² = 16 and b² = 9. Hence a = 4 and b = 3.

For such a hyperbola, e = √(1 + b²/a²).

Therefore, e = √(1 + 9/16) = √(25/16) = 5/4.

Hence the eccentricity is 5/4. (Option B)

Let the equation of tangent drawn from (0,0) to the circle be y=mx. Then, p = a ⇒ 7m+1/(m²+1)^1/2= 5
⇒24m² + 14m−24=0
⇒12m² + 7m−12=0
⇒m1m² = −12/12 =−1
∴ Required angle = π/2