Find the equation of the line through the point (3, 4) which cut

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsLong Answer Type

341. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 over 27 of the volume of the sphere.
Or
Find the volume of the largest cone that can be inscribed in a sphere of radius r.

97 Views

342.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan to the power of negative 1 end exponent square root of 2.

77 Views

343.

For a given current surface area of right circular cone when the volume is maximum. Prove that the semi-vertical angle is straight theta where sin space straight theta space equals space fraction numerator 1 over denominator square root of 3 end fraction.

84 Views

 Multiple Choice QuestionsShort Answer Type

344.

Prove that a conical tent of given capacity will require the least amount of convas when the height is square root of 2 times the radius of the base.

103 Views

Advertisement
345.

Show that the right circular cone of least curved surface and given volume has an altitude equal to an altitude equal to square root of 2 times the radius of the base.

121 Views

 Multiple Choice QuestionsLong Answer Type

346.

Find the altitude of a right circular cone of maximum curved surface which can be inscribed in a sphere of radius r.

299 Views

Advertisement

347.

Find the equation of the line through the point (3, 4) which cuts from the first quadrant a triangle of minimum area.


The equation of any line through P (3, 4) is
   straight y minus 4 space equals space straight m left parenthesis straight x minus 3 right parenthesis space space space space space space space... left parenthesis 1 right parenthesis
where m is slope of line
   Put y = 0 in (1)
therefore space space space space space minus 4 space equals space mx space minus space 3 straight m space space space space space rightwards double arrow space space space mx space equals space 3 straight m minus 4
rightwards double arrow space space space space space space straight x space equals space 3 minus 4 over straight m
therefore space space space space space space line space left parenthesis 1 right parenthesis space meets space straight x minus axis
space space space space space space space space space space space space space in space straight A space open parentheses 3 minus 4 over straight m comma space 0 close parentheses
Again space put space straight x space equals space 0 space in space left parenthesis 1 right parenthesis
therefore space space space space space space space space straight y minus 4 space equals space minus 3 space straight m space space space rightwards double arrow space space space space straight y space equals space minus 3 straight m plus 4
therefore space space space space space line space left parenthesis 1 right parenthesis space meets space straight y minus axis space in space straight B left parenthesis 0 comma space minus 3 straight m space plus 4 right parenthesis

Let ∆ denote the area of ∆OAB
  therefore space space space space increment space equals space 1 half OA. space space OB space equals space 1 half open parentheses 3 minus 4 over straight m close parentheses space left parenthesis negative 3 straight m space plus space 4 right parenthesis
space space space space space space space space space space space space equals space 1 half open parentheses negative 9 straight m space plus 12 space plus space 12 space minus 16 over straight m close parentheses space equals space minus 9 over 2 straight m minus 8 over straight m
space space space space space fraction numerator straight d increment over denominator dm end fraction space equals space minus 9 over 2 plus 8 over straight m squared
space space space space space space space space fraction numerator straight d squared increment over denominator dm squared end fraction space equals space minus 16 over straight m cubed

For increment to be maximum or minimum,
          space fraction numerator straight d increment over denominator dm end fraction space equals space 0 space space space space or space space minus 9 over 2 plus 8 over straight m squared space equals space 0 space space space space space space rightwards double arrow space space space minus 9 straight m squared plus 16 space equals space 0

rightwards double arrow space space space straight m squared space equals space 16 over 9 space space space rightwards double arrow space space space space straight m space equals space minus 4 over 3 comma space space 4 over 3

When space straight m space equals space minus 4 over 3 comma space fraction numerator straight d squared increment over denominator dm squared end fraction greater than 0
therefore space space space space space space space increment space space space is space minimum space when space straight m space equals space minus 4 over 3
When space space space straight m space equals space 4 over 3 comma space space space fraction numerator straight d squared increment over denominator dm squared end fraction less than 0
therefore space space space space space space straight A space is space maximum space when space straight m space equals space 4 over 3
therefore space space space space space space space increment space space is space minimum space when space straight m space equals space minus 4 over 3
Putting space straight m space equals space minus 4 over 3 space in space left parenthesis 1 right parenthesis comma space we space get comma
space space space space space space space space space space straight y minus 4 space equals space minus 4 over 3 left parenthesis straight x minus 3 right parenthesis space space space space space space space space space or space space space space 3 straight y minus 12 space equals space minus 4 straight x plus 12
or space space space 4 straight x plus 3 straight y space equals space 24 comma
space
which is required equation of line.


73 Views

Advertisement
348.

Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.

81 Views

Advertisement
349. Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.
377 Views

350.

Prove that the perimeter of a right-angled triangle of given hypotenuse equal to 5 cm is maximum when the triangle is isosceles.

90 Views

Advertisement