JEE Main 2018 (Online) 16th April Morning Slot | Hyperbola Question 64 | Mathematics

The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$$ (k is any non-zero real parameter), is :

an ellipse whose eccentricity is $${1 \over {\sqrt 3 }}.$$

an ellipse with length of its major axis $$8\sqrt 2 .$$

a hyperbola whose eccentricity is $$\sqrt 3 .$$

a hyperbola with length of its transverse axis $$8\sqrt 2 .$$

JEE Main 2018 (Offline)

MCQ (Single Correct Answer)

-1

Out of Syllabus

Tangents are drawn to the hyperbola 4x² - y² = 36 at the points P and Q.

If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $$\Delta $$PTQ is :

$$36\sqrt 5 $$

$$45\sqrt 5 $$

$$54\sqrt 3 $$

$$60\sqrt 3 $$

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

-1

Out of Syllabus

A normal to the hyperbola, 4x² $$-$$ 9y² = 36 meets the co-ordinate axes $$x$$ and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is :

4x² + 9y² = 121

9x² + 4y² = 169

4x² $$-$$ 9y² = 121

9x² $$-$$ 4y² = 169

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

-1

If the tangents drawn to the hyperbola 4y² = x² + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is :

x² $$-$$ 4y² + 16x²y² = 0

x² $$-$$ 4y² $$-$$ 16x²y² = 0

4x² $$-$$ y² + 16x²y² = 0

4x² $$-$$ y² $$-$$ 16x²y² = 0

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