Answer:
[tex]\qquad\qquad \: \: \boxed{ \bf{ \:\:(b)\: \: (1, \: 2) \: \: }} \\ \\ [/tex]
Step-by-step explanation:
Let assume that (h, k) be the point on the curve [tex]y^2 [/tex] = 4x which is nearest to the point (2, 1).
Let assume that coordinates (h, k) and (2, 1) represented as P and A respectively.
As P(h, k) lies on the curve [tex]y^2 [/tex] = 4x.
[tex]\bf\implies \: {k}^{2} = 4h \: \: \bf\implies h = \dfrac{ {k}^{2} }{4} - - - (1) \\ \\ [/tex]
Now,
[tex]\sf \: AP = \sqrt{ {(h - 2)}^{2} + {(k - 1)}^{2} } \\ \\ [/tex]
[tex]\sf \: {AP}^{2} = { {(h - 2)}^{2} + {(k - 1)}^{2} } \\ \\ [/tex]
[tex]\sf \: {AP}^{2} = {\left(\dfrac{ {k}^{2} }{4} - 2\right)}^{2} + {(k - 1)}^{2} \\ \\ [/tex]
[tex]\sf \: f(k)= {\left(\dfrac{ {k}^{2} }{4} - 2\right)}^{2} + {(k - 1)}^{2} \\ \\ [/tex]
[tex]\sf \: f(k)= \dfrac{ {k}^{4} }{16} + 4 - {k}^{2} + {k}^{2} + 1 - 2k \\ \\ [/tex]
[tex]\sf \: f(k)= \dfrac{ {k}^{4} }{16} + 5 - 2k \\ \\ [/tex]
On differentiating both sides w. r. t. k, we get
[tex]\sf \: f'(k)= \dfrac{ 4{k}^{3} }{16} - 2 \\ \\ [/tex]
[tex]\sf \: f'(k)= \dfrac{{k}^{3} }{4} - 2 - - - (i)\\ \\ [/tex]
For maxima or minima,
[tex]\sf \: f'(k)= 0 \\ \\ [/tex]
[tex]\sf \: \dfrac{{k}^{3} }{4} - 2= 0 \\ \\ [/tex]
[tex]\sf \: \dfrac{{k}^{3} }{4} = 2 \\ \\ [/tex]
[tex]\sf \: {k}^{3} = 8 \\ \\ [/tex]
[tex]\bf\implies \: k = 2 - - - (2) \\ \\ [/tex]
Now, from equation (i) we have
[tex]\sf \: f'(k)= \dfrac{{k}^{3} }{4} - 2 \\ \\ [/tex]
[tex]\sf \: f''(k)= \dfrac{3{k}^{2} }{4} \\ \\ [/tex]
On substituting k = 2, we get
[tex]\sf \: f''(2)= \dfrac{3{(2)}^{2} }{4} = 3 > 0 \\ \\ [/tex]
[tex]\bf\implies \: \: f(k) \: is \: minimum \\ \\ [/tex]
[tex]\bf\implies \: \: AP \: is \: minimum \\ \\ [/tex]
On substituting k = 2 in equation (1), we get
[tex]\sf \: h = \dfrac{ {2}^{2} }{4} \\ \\ [/tex]
[tex]\sf \: h = \dfrac{ 4 }{4} \\ \\ [/tex]
[tex]\bf\implies \: \: h = 1 \\ \\ [/tex]
Hence,
[tex]\bf\implies \: \: \boxed{ \bf{ \:Coordinates \: of \: P \:be \: (1, \: 2) \: \: }} \\ \\ [/tex]
Thus,
(1, 2) be the point on the curve [tex]y^2 [/tex] = 4x which is nearest to the point (2, 1).
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Answers & Comments
Answer:
[tex]\qquad\qquad \: \: \boxed{ \bf{ \:\:(b)\: \: (1, \: 2) \: \: }} \\ \\ [/tex]
Step-by-step explanation:
Let assume that (h, k) be the point on the curve [tex]y^2 [/tex] = 4x which is nearest to the point (2, 1).
Let assume that coordinates (h, k) and (2, 1) represented as P and A respectively.
As P(h, k) lies on the curve [tex]y^2 [/tex] = 4x.
[tex]\bf\implies \: {k}^{2} = 4h \: \: \bf\implies h = \dfrac{ {k}^{2} }{4} - - - (1) \\ \\ [/tex]
Now,
[tex]\sf \: AP = \sqrt{ {(h - 2)}^{2} + {(k - 1)}^{2} } \\ \\ [/tex]
[tex]\sf \: {AP}^{2} = { {(h - 2)}^{2} + {(k - 1)}^{2} } \\ \\ [/tex]
[tex]\sf \: {AP}^{2} = {\left(\dfrac{ {k}^{2} }{4} - 2\right)}^{2} + {(k - 1)}^{2} \\ \\ [/tex]
[tex]\sf \: f(k)= {\left(\dfrac{ {k}^{2} }{4} - 2\right)}^{2} + {(k - 1)}^{2} \\ \\ [/tex]
[tex]\sf \: f(k)= \dfrac{ {k}^{4} }{16} + 4 - {k}^{2} + {k}^{2} + 1 - 2k \\ \\ [/tex]
[tex]\sf \: f(k)= \dfrac{ {k}^{4} }{16} + 5 - 2k \\ \\ [/tex]
On differentiating both sides w. r. t. k, we get
[tex]\sf \: f'(k)= \dfrac{ 4{k}^{3} }{16} - 2 \\ \\ [/tex]
[tex]\sf \: f'(k)= \dfrac{{k}^{3} }{4} - 2 - - - (i)\\ \\ [/tex]
For maxima or minima,
[tex]\sf \: f'(k)= 0 \\ \\ [/tex]
[tex]\sf \: \dfrac{{k}^{3} }{4} - 2= 0 \\ \\ [/tex]
[tex]\sf \: \dfrac{{k}^{3} }{4} = 2 \\ \\ [/tex]
[tex]\sf \: {k}^{3} = 8 \\ \\ [/tex]
[tex]\bf\implies \: k = 2 - - - (2) \\ \\ [/tex]
Now, from equation (i) we have
[tex]\sf \: f'(k)= \dfrac{{k}^{3} }{4} - 2 \\ \\ [/tex]
[tex]\sf \: f''(k)= \dfrac{3{k}^{2} }{4} \\ \\ [/tex]
On substituting k = 2, we get
[tex]\sf \: f''(2)= \dfrac{3{(2)}^{2} }{4} = 3 > 0 \\ \\ [/tex]
[tex]\bf\implies \: \: f(k) \: is \: minimum \\ \\ [/tex]
[tex]\bf\implies \: \: AP \: is \: minimum \\ \\ [/tex]
On substituting k = 2 in equation (1), we get
[tex]\sf \: h = \dfrac{ {2}^{2} }{4} \\ \\ [/tex]
[tex]\sf \: h = \dfrac{ 4 }{4} \\ \\ [/tex]
[tex]\bf\implies \: \: h = 1 \\ \\ [/tex]
Hence,
[tex]\bf\implies \: \: \boxed{ \bf{ \:Coordinates \: of \: P \:be \: (1, \: 2) \: \: }} \\ \\ [/tex]
Thus,
(1, 2) be the point on the curve [tex]y^2 [/tex] = 4x which is nearest to the point (2, 1).