IIT-JEE 2003 Screening | Straight Lines and Pair of Straight Lines Question 16 | Mathematics

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.75

Orthocentre of triangle with vertices $$\left( {0,0} \right),\left( {3,4} \right)$$ and $$\left( {4,0} \right)$$ is

$$\,\,\left( {3,{5 \over 4}} \right)$$

$$\left( {3,12} \right)$$

$$\left( {3,{3 \over 4}} \right)$$

$$\left( {3,9} \right)$$

IIT-JEE 2002

MCQ (Single Correct Answer)

-1

The pair of lines represented by
$$3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0$$ are perpendicular to each other for

two values of $$a$$

$$\forall \,a$$

for one values of $$a$$

for no values of $$a$$

IIT-JEE 2002

MCQ (Single Correct Answer)

-1

If the pair of lines $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$ intersect on the $$y$$ axis then

$$2fgh = b{g^2} + c{h^2}$$

$$b{g^2} \ne c{h^2}$$

$$\,abc = 2fgh$$

none of these

IIT-JEE 2002

MCQ (Single Correct Answer)

-1

Locus of mid point of the portion between the axes of $$x$$ $$\cos \alpha + y\sin \alpha = p$$ where $$p$$ is constant is

$${x^2} + {y^2} = {4 \over {{p^2}}}\,\,\,$$

$${x^2} + {y^2} = 4{p^2}$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {2 \over {{p^2}}}$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}$$

JEE Advanced Subjects