JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Medium

1 of 1

Application +4 | -1

The abscissa of the point on the curve

9 y^2

=x^3

, the normal at which cuts off equal intercepts on the coordinate axes is

View answer and explanation

Correct option: B

Option B (4) is correct because the abscissa of the point is 4, which is found by setting the intercepts of the normal equal and solving for x.

More Questions from PROPERTIES OF TRIANGLES

Practice All Chapter MCQs →

{p_1},{p_2},{p_3}

\Delta \;ABC

A,B,C

\Delta

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

5k,\;6k,\;5k

k

\Delta \,ABC

a = \left( {b - c} \right)\sec \theta

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

\Delta \;ABC,\frac{{b + c}}{a}

\Delta = {a^2} - {\left( {b - c} \right)^2}

\Delta

\Delta \;ABC

\tan A

A

2 s