WB JEE 2020 | Straight Lines and Pair of Straight Lines Question 42 | Mathematics

WB JEE 2020

MCQ (Single Correct Answer)

-0.5

A line cuts the X-axis at A(7, 0) and the Y-axis at B(0, $$ - $$5). A variable line PQ is drawn perpendicular to AB cutting the X-axis at P(a, 0) and the Y-axis at Q(0, b). If AQ and BP intersect at R, the locus of R is

$${x^2} + {y^2} + 7x + 5y = 0$$

$${x^2} + {y^2} + 7x - 5y = 0$$

$${x^2} + {y^2} - 7x + 5y = 0$$

$${x^2} + {y^2} - 7x - 5y = 0$$

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

A variable line passes through a fixed point $$({x_1},{y_1})$$ and meets the axes at A and B. If the rectangle OAPB be completed, the locus of P is, (O being the origin of the system of axes).

$${(y - {y_1})^2} = 4(x - {x_1})$$

$${{{x_1}} \over x} + {{{y_1}} \over y} = 1$$

$${x^2} + {y^2} = x_1^2 + y_{\kern 1pt} ^2$$

$${{{x^2}} \over {2x_1^2}} + {{{y^2}} \over {y_1^2}} = 1$$

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

A straight line through the point (3, $$-$$2) is inclined at an angle 60$$^\circ$$ to the line $$\sqrt 3 x + y = 1$$. If it intersects the X-axis, then its equation will be

$$y + x\sqrt 3 + 2 + 3\sqrt 3 = 0$$

$$y - x\sqrt 3 + 2 + 3\sqrt 3 = 0$$

$$y - x\sqrt 3 - 2 - 2\sqrt 3 = 0$$

$$x - x\sqrt 3 + 2 - 3\sqrt 3 = 0$$

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

A variable line passes through the fixed point $$(\alpha ,\beta )$$. The locus of the foot of the perpendicular from the origin on the line is

$${x^2} + {y^2} - \alpha x - \beta y = 0$$

$${x^2} - {y^2} + 2\alpha x + 2\beta y = 0$$

$$\alpha x + \beta y \pm \sqrt {({\alpha ^2} + {\beta ^2})} = 0$$

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

PREVIOUS NEXT

WB JEE Subjects